Find the volume of this triangular prism.

Answer:

42cm3

Step-by-step explanation:

v = 0.5× b × h × l

= 0.5 × 4 × 3 × 7

= 42cm3

Answer:

131.95yd squared

Step-by-step explanation:

A=2πrh+ 2 r(2)=2*π*3*4+2*π*3(2)~131.94689yd(2)

Answer:

Yay fun day :DDDDDD

Step-by-step explanation:

Answer:the answer is solid B

Step-by-step explanation:

Oh your on edmentum I hate it there is so much work but I can help I'm a math master I'm top 3 smartest at math in my whole school.

Answer:

not a complete question..

Step-by-step explanation:

Answer:

The experimental probability of rolling anumber less than 3 is 3/20

Step-by-step explanation:

16+14=30

30/200

=3/20

Answer:

7/50

Step-by-step explanation:

Answer:

models

Step-by-step explanation:

Answer: B.) Models

Edge

The types of transformation in this problem is given as follows:

Vertical and horizontal translation.

The four translation rules are defined as follows:

The translations for this problem are given as follows:

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Answer:

The neighborhood the home is located in

The factors which does not affect the mortgage payment is the neighborhood the home is located in.

The correct option is D.

We are aware of this;

A mortgage is a loan when the borrower's property is used as security. The mortgage payment is determined by the cost of the property, the interest rate, the down payment, the length of the loan, taxes, and various insurances like homeowners insurance, among other factors.

Hence, It's not depend on ''The neighborhood the home is located in.''

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Answer: January 1, 2025

Step-by-step explanation:

The debt is due to be paid back in 10 years.

If the first payment was in 2016, the last payment should therefore be:

= 2016 + 9 years

= 2025

We used 9 years because 2016 was the first year of payment so the remaining years would be 9 years.

As the first payment was on January 1, 2016, the last payment would have to be on the same date in 2025 which is:

= January 1, 2025

Shape a = 6 x 4 x 3 = 72

Shape b = 4 x 4 x 4 = 64

72 + 64 = 136

A

Answer:

A

Step-by-step explanation:

(a) If there are 4 candidate,  the smallest number of votes that a plurality candidate could have is 33.

(b) If there are 8 candidate,  the smallest number of votes that a plurality candidate could have is 17.

The smallest number of votes that a plurality candidate could have is calculated as follows;

(a) If there are 4 candidate, the number of votes for each candidate;

= 129 / 4

= 32.25

The least number of votes for the plurality candidate = 33

(b) If there are 8 candidate, the number of votes for each candidate;

= 129 / 8

= 16.125

The least number of votes for the plurality candidate = 17

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Answer:

(X+6)(x-5) + (X-5)(X+6)

Answer:

[tex]2x ^{2} - 22x + 60[/tex]

Answer:

-10x+30

Step-by-step explanation:

just got it right on edg 2021 :)

None of these are correct

Given that a hockey tournament consists of 16 teams. In the first round, every team is randomly assigned to one of 8 games (2 teams per game). We are supposed to find the probability that all three Alberta teams are randomly assigned to different games. Let A be the event of assigning all three Alberta teams to different games.

Then the number of ways to select 3 teams from 16 teams is $\ dbinom {16}{3}$, the number of ways to assign 3 teams to different games is $8\times7\times6$, and the number of ways to assign the remaining 13 teams to games is $(13!) / (2^6\times6!)$.The probability of event A is given by;$$

P(A) = \frac{\text{number of ways to assign 3 teams from Alberta to different games}}{\text{number of ways to assign all teams to games}} = \frac{8\times7\times6 \times (13!) / (2^6\times6!)}{\dbinom{16}{3} \times (14!) / (2^7\times7!)}
$$Simplifying the above expression,$$
P(A) = \frac{8\times7\times6 \times 13! \times 2}{\dbinom{16}{3} \times 14!} = \frac{8\times7\times6 \times 2}{\dbinom {16}{3}} = \frac{336}{560} = \frac{3}{5}

Therefore, the probability that all three Alberta teams are randomly assigned to different games is $\frac{3}{5}$.

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To find the second derivative of F(x) = [tex]\int\limits^0_x {e^t}^{2} } } \, dx[/tex] dt for x ∈ (0, 1), we can use the fundamental theorem of calculus and apply the chain rule twice. The second derivative is given by F''(x) = [tex]2e^{x^{2} } (1+2x^{2} )[/tex]

To find F''(x), we differentiate F'(x) first. Using the fundamental theorem of calculus, we have F'(x) = [tex]e^{x^{2} }[/tex]. Applying the chain rule, we obtain d/dx([tex]e^{x^{2} }[/tex]) = [tex]2xe^{x^{2} }[/tex].

Now, to find F''(x), we differentiate F'(x) with respect to x. Applying the chain rule again, we have d/dx([tex]2xe^{x^{2} }[/tex]) = [tex]2e^{x^{2} }[/tex] + [tex]4x^{2} e^{x^{2} }[/tex]. Simplifying this expression, we get F''(x) = 2[tex]e^{x^{2} }[/tex](1 + [tex]2e^{x^{2} }[/tex]).

Therefore, the second derivative of F(x) is given by F''(x) = 2[tex]e^{x^{2} }[/tex](1 + 2[tex]{x^{2} }[/tex]) for x ∈ (0, 1). This result shows that the second derivative is always positive for x ∈ (0, 1), indicating that the function is concave up within that interval.

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Answer:

x axis

Step-by-step explanation:

[tex]\Large\boxed{\tt Answer:~A:~y=-3x+5}[/tex]

All we have to do in order to find the equation that is equivalent to 6x + 2y = 10 is to solve for y.

Step 1: Keep y on one side of the equation and move the rest to the other.

[tex]\tt 6x + 2y = 10\\2y = -6x + 10[/tex]

Step 2: Divide all terms by 2 to isolate y.

[tex]\tt 2y \div 2=y\\-6x \div 2 = -3x\\10 \div 2 =5\\\\y=-3x+5[/tex]

To calculate[tex]2^2873686243768478237864767208[/tex] mod 101 using Fermat's little theorem, we can simplify the exponent by taking it modulo 100, since 100 is the Euler's totient function value of 101. Therefore,[tex]2^2873686243768478237864767208[/tex] mod 101 is equal to 57.

Fermat's little theorem states that if p is a prime number and a is any integer not divisible by p, then a^(p-1) ≡ 1 (mod p). In this case, p = 101, and we need to find[tex]2^2873686243768478237864767208[/tex]mod 101.

First, we simplify the exponent by taking it modulo 100, since 100 is the Euler's totient function value of 101. The exponent 2873686243768478237864767208 is congruent to 8 modulo 100. So, we need to calculate 2^8 mod 101. Applying Fermat's little theorem, we know that 2^(101-1) ≡ 1 (mod 101), since 101 is prime. Therefore, 2^100 ≡ 1 (mod 101).

We can express [tex]2^8[/tex] in terms of 2^100 as [tex](2^100)^0.08[/tex]. Simplifying this, we get [tex](2^100)^0.08 ≡ 1^0.08[/tex]≡ 1 (mod 101).

Thus, we conclude that[tex]2^8[/tex] ≡ 1 (mod 101), and therefore 2^2873686243768478237864767208 ≡ [tex]2^8[/tex] (mod 101).

Finally, evaluating [tex]2^8[/tex] mod 101, we find that [tex]2^8[/tex] ≡ 57 (mod 101).

Therefore,[tex]2^2873686243768478237864767208[/tex] mod 101 is equal to 57.

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[tex]\frac{-3 - (-4)}{1 - (-5)}[/tex]

= [tex]\frac{-3 + 4}{1 + 5}[/tex]

= 1/6

Answer:

267/500

Step-by-step explanation:

0.534 = 534 / 1000

Simplify to 267/500

Step-by-step explanation:

537/1000

this is the correct answer

Answer:

1/2 x 1/2 = 1/4

Step-by-step explanation:

P(H,H) = 1/2 x 1/2 = 1/4

P(T,T) = 1/2 x 1/2 = 1/4

P(H,T) = 1/2 x 1/2 = 1/4

P(T,H) = 1/2 x 1/2 = 1/4

Answer:

37.36 meters are equal to 3,736 centimeters.

Step-by-step explanation:

We have that 1 meter is equal to 100 centimeters, so if we want to convert 3,736 cm to m we need to divide by 100:

[tex] x = 3,736 cm*\frac{1 m}{100 cm} = 3,736 cm*\frac{1 m}{10^{2} cm} [/tex]          

When we divide a number by a power of 10, we move the decimal point to the left as many places as the power indicates.      

Since we are dividing by 10², we need to move the decimal point two places to the left, as follows:            

[tex] x = 3,736 cm*\frac{1 m}{10^{2} cm} = 37.36 m [/tex]      

Hence, 37.36 meters are equal to 3,736 centimeters.

I hope it helps you!                  

Mathematical induction can be used to prove that for every non-negative odd integer n, the expression [tex]24 / (2^{(3n+1)}+1+1) * (n^2+1)[/tex] holds true.

To prove the statement using mathematical induction, we need to follow two steps: the base case and the induction step.

First, we verify if the statement holds true for the base case, which is typically the smallest value of n. In this case, let's consider n = 0. Plugging in n = 0 into the expression, we get [tex]24 / (2^{(3*0+1)}+1+1) * (0^2+1)[/tex]. Simplifying, we have 24 / (2+1+1) * 1, which equals 24 / 4 * 1, resulting in 6. Therefore, the statement holds true for n = 0.

Next, we assume that the statement is true for some arbitrary odd integer k, and we will prove that it holds true for k+2. Assume that [tex]24 / (2^{(3k+1)}+1+1) * (k^2+1)[/tex] holds true.

Now, we substitute k+2 into the expression and aim to show that it holds true for k+2 as well. We have [tex]24 / (2^{(3(k+2)+1)}+1+1) * ((k+2)^2+1)[/tex]. Simplifying the expression, we get [tex]24 / (2^{(3k+7)}+1+1) * (k^2 + 4k + 5)[/tex].

We can manipulate the equation further to demonstrate that it is equal to the assumed expression for k. By performing algebraic manipulations and simplifications, we can equate the expressions and conclude that the statement holds true for k+2.

Since we have verified the base case and shown that the statement holds true for k+2 when it holds true for k, we can conclude that the statement is true for every non-negative odd integer n.

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Answer:

5 batches

Step-by-step explanation:

1/6 oatmeal  can make 1 batch, so 5/6 makes 5 batches

Answer:

38 ft ^2

Step-by-step explanation:

8x 3 = 24

1/2(4)(3) = 6

1/2(4)(4) = 8

24+6+8 = 38ft^2

Answer:

41.7feet

Step-by-step explanation:

From the question we are given the following

angle of depression = 50°

Distance of the pole from the base of the feet = 35feet (Adjacent)

Required

height of the school (opposite)

Using the SOH CAH TOA identity

Tan theta = opp/adj

Tan 50 = H/35

H = 35tan 50

H = 35(1.1918)

H = 41.7feet

Hence the height of the school is 41.7feet

The following can cause OLS estimators to be biased: Omitting an important independent variable.

Multicollinearity. Heteroskedasticity, Including an irrelevant variable.

The following does not cause the usual OLS t statistics to be invalid (that is, to have t distributions under H0): The error term non-normally distributed

OLS (ordinary least squares) estimates are typically unbiased when calculated.

However, the following problems may cause OLS estimates to be biased:

Omitting an important independent variable: When an important independent variable is omitted from the regression equation, the OLS estimate of the effect of one variable on the dependent variable is biased.

In particular, the estimate of the effect of the variable that is omitted is influenced by the remaining variables' presence in the equation.

Multicollinearity: When the independent variables in a multiple regression model are strongly related, multicollinearity exists.

When there is multicollinearity in a model, the estimated slope coefficients are frequently biased, making them difficult to interpret.

In this scenario, small changes in the data may cause substantial changes in the estimated coefficients.

As a result, the usual tests of hypothesis may fail to produce reliable inferences.

Heteroskedasticity: In the population, heteroskedasticity exists when the variance of the error term is not constant across observations.

Heteroskedasticity can induce OLS estimates' variance to be biased, even if the estimates are unbiased themselves.

When there is heteroskedasticity, the OLS estimates are no longer BLUE (best linear unbiased estimator).

Including irrelevant variable: When an irrelevant variable is included in a regression equation, the OLS estimates of the other variables' effects are biased, and the estimates' standard errors are larger than necessary.

The error term non-normally distributed: When the error term in a regression equation is non-normally distributed, the distribution of the OLS estimates is also non-normal.

However, this does not affect the distribution of the t statistics under H0.

The reason for this is that, even if the error term is non-normally distributed, the sample mean converges to the population mean, according to the central limit theorem.

Furthermore, the standard error of the mean is unaffected by the distribution of the error term, as long as the sample size is large enough.

As a result, the t statistics can be trusted to be asymptotically normally distributed under H0.

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Answer:

Step-by-step explanation:

h

Answer:

51 and 261

Step-by-step explanation:

Let the numbers be x and y

x + y =312

x - y =210

2y=102

y=51

x=312-51=261