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3 years ago

PLSSSSS HELPPPPPP MEH ;^; What is the lateral surface area of the triangular prism?

Mathematics

2 answers:

Zarrin [17]3 years ago

8 0

Answer:

I think it's 56.4

Step-by-step explanation:

if not I'm sorry

Klio2033 [76]3 years ago

5 0

GURL i have the same question *cries in math *

You might be interested in

The area of a regular hexagon inscribed in a circle with radius 2 is . Round your answer to the nearest tenth.

Verdich [7]

Answer:

10.4 to nearest tenth.

Step-by-step explanation:

If we draw lines from the centre of the circle to each vertex of the hexagon we get 6 equilateral triangles of side length 2.

Altitude of each triangle = sqrt(2^2 - 1^2)

= sqrt3.

So the area of 1 triangle = 1/2 * 2 * sqrt3

= sqrt3

Therefore the area of the hexagon = 6 * sqrt3

= 6sqrt3

= 10.3923

6 0

3 years ago

Location is known to affect the number, of a particular item, sold by HEB Pantry. Two different locations, A and B, are selected

12345 [234]

Answer:

d) [-16.03,-3.97]

$-16.03 \leq \mu_A -\mu_B \leq -3.97$ .

Step-by-step explanation:

Notation and previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$n_A=18$ represent the sample of A

$n_B =13$ represent the sample of B

$\bar x_A =39$ represent the mean sample for A

$\bar x_B =49$ represent the mean sample for B

$s_A =8$ represent the sample deviation for A

$s_B =4$ represent the sample deviation for B

$\alpha=0.01$ represent the significance level

Confidence =99% or 0.99

The confidence interval for the difference of means is given by the following formula:

$(\bar X_A -\bar X_B) \pm t_{\alpha/2}\sqrt{(\frac{s^2_A}{n_A}+\frac{s^2_B}{n_B})}$ (1)

The point of estimate for $\mu_A -\mu_B$ is just given by:

$\bar X_A -\bar X_B =39-49=-10$

The appropiate degrees of freedom are $df=n_1+ n_2 -2=18+13-2=29$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,29)".And we see that $t_{\alpha/2}=2.756$

The standard error is given by the following formula:

$SE=\sqrt{(\frac{s^2_A}{n_A}+\frac{s^2_B}{n_B})}$

And replacing we have:

$SE=\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=2.188$

Confidence interval

Now we have everything in order to replace into formula (1):

$-10-2.756\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=-16.03$

$-10+2.756\sqrt{(\frac{8^2}{18}+\frac{4^2}{13})}=-3.97$

So on this case the 99% confidence interval for the differences of means would be given by $-16.03 \leq \mu_A -\mu_B \leq -3.97$ .

d) [-16.03,-3.97]

7 0

4 years ago

Given a square with two vertices of one side located at (-5. -3) and (-5, 12), in square units what is its area?

givi [52]

Answer:

4. 225

Step-by-step explanation:

The distance between points (-5, -3) and (-5, 12) is the length of the side of the square. Since both points have an x-coordinate of -3, they lie on the same vertical line. The distance between them is |12 - (-3)| = |12 + 3| = |15| = 15. The side measures 15 units.

area = side^2

area = 15^2 = 225

4 0

4 years ago

Read 2 more answers

9(2+2p) show distributive property

balu736 [363]

Answer:

9(2+2p)

9 x2 + 9 x 2p

18 +18p

6 0

3 years ago

How do you write 5.9 in expanded form

tresset_1 [31]

Answer:
(5 * 1) + (9 * 0.1).

Explanation:
To write a number in expanded form you have each digit multiplied by its numbers place and add each digit together. 5 is in the ones place so it becomes (5 * 1).
9 is in the tenths place so it becomes (9 * 0.1).
add them together to get (5 * 1) + (9 * 0.1)

3 0

3 years ago

Read 2 more answers