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2 years ago

Find the total surface area of the rectangular pyramid.

Mathematics

2 answers:

pantera1 [17]2 years ago

5 0

Answer:

I can't really understand the wording but..

Step-by-step explanation

The formula for surface area of a rectangular prism is A=2(wl+hl+hw)

so yeah enjoy freedom for the rest of the day

VashaNatasha [74]2 years ago

4 0

12 because 12 is never wrong and 12 is an amazing number

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What type of transformation is used when the line y = vx is transformed into y= 5x?

dedylja [7]

Answer:

It gains a slope of 5 over 1

Step-by-step explanation:

4 0

3 years ago

A farmer needs to ship 71 pumpkins to a grocery store if each crates can hold 19 pumpkins how many crates Will the farmer need ?

Lesechka [4]

Answer:

The farmer will need 7 crates.

Step-by-step explanation:

It is given that there are 71 pumpkins in total.

And each crate holds 19 pumpkins.

Let the number of crates required be "n".

The total number of pumpkins can be calculated by multiplying number of pumpkins in each crate with total number of crates.

Thus, the equation is

$19(n) = 71$

$n = \frac{71}{19} = 7$

Thus, the farmer needs 7 crates to hold total of 71 pumpkins.

3 0

3 years ago

Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r

ZanzabumX [31]

Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean weight = $\frac{\sum X}{n}$ = 21.88 ounces

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 0.201 ounces

n = sample of boxes = 12

$\mu$ = population mean weight

Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.796 < $t_1_1$ < 1.796) = 0.90 {As the critical value of t at 11 degrees of

freedom are -1.796 & 1.796 with P = 5%}

P(-1.796 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.796) = 0.90

P( $-1.796 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $21.88-1.796 \times {\frac{0.201}{\sqrt{12} } }$ , $21.88+1.796 \times {\frac{0.201}{\sqrt{12} } }$ ]

= [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

8 0

3 years ago

At the start of the month, Suzanne's clothing store website had 7,278 hits. At the end of the month, it had 50,165 hits. If each

anyanavicka [17]

Answer:

Option A. 64,500

Explanation:

1. First, calculate the number of hits had during the month, by difference:

50,165 hits - 7,278 hits = 42,887 hits

2. Round the result to obtain an approximation = 43,000 hits

3. Multiply by the time that each lasted:

43,000 hits × 1.5 min/hit = 64,500 minutes.

Hence, the answer is the option A.

3 0

3 years ago

V = Bh for h. A. h = BV B. h=B/V C. h=V/B D. h = V – B

Darina [25.2K]

C. h=V/B to solve for this you need to get the variable h by itself and since h is being multiplied to B you use the opposite operation of division by divide both sides by B. Then you have h solved for h=V/B

5 0

3 years ago