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3 years ago

This question below.

Mathematics

1 answer:

pochemuha3 years ago

6 0

I can’t really see what it says but it it’s total surface area it’s 9044.3 but it it’s lateral surface are it’s 7840

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Is 1 a probability?

kaheart [24]

Yes one is a probability because it can be one out of 4 or something
Good luck hope this helps

3 0

3 years ago

from a balloon 1191 feet high, the angle of depression to the ranger headquarters is 72°54'. how far is the headquarters from a

lara [203]

Since angle of depression = 72°54 minutes.

60 minutes = 1°

54 minutes = 54/60 = 0.9°

Therefore 72°54 minutes = 72° + 0.9° = 72.9°

Since angle of depression = 72.9°, the reverse angle of elevation = 72.9°

(Picture representation of this using right angled triangle is needed)

tan 72.9° = Opposite height / Horizontal distance

tan 72.9° = 1191 / x

x = 1191 / tan 72.9 Use a calculator.

x ≈1191 / 3.2506

x ≈ 366.39

Horizontal distance ≈ 366.39 feet.

Hope this helped.

6 0

3 years ago

A wholesaler buys a pair of sunglasses for $5.50 each. He then marks up the wholesale cost 80% to create a retail price for his

lorasvet [3.4K]

.80+5.50=6.30

6.30+.07=6.37

so 6.37 would be it

8 0

3 years ago

The sum of a number and 35 is 100 create an equation

Gnesinka [82]

Let's called this number x and sum it with 35

x+35

We know this sum is equal to 100, let's equal it.

x+35=100

$\boxed { x+35=100 }$

3 0

3 years ago

Read 2 more answers

A study is performed in San Antonio to determine whether the average weekly grocery bill per five-person family in the town is s

bogdanovich [222]

Answer:

No, the sample evidence is not statistically significant (P-value = 0.125).

To reject the null hypothesis, the P-value has to be smaller than the significance level, so the significance level to reject the null hypothesis has to be 0.125 or higher.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the average weekly grocery bill per five-person family in San Antonio is significantly different from the national average.

Then, the null and alternative hypothesis are:

$H_0: \mu=131\\\\H_a:\mu\neq 131$

The significance level is 0.05.

The sample has a size n=50.

The sample mean is M=133.474.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.193.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11.193}{\sqrt{50}}=1.583$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{133.474-131}{1.583}=\dfrac{2.474}{1.583}=1.563$

The degrees of freedom for this sample size are:

$df=n-1=50-1=49$

This test is a two-tailed test, with 49 degrees of freedom and t=1.563, so the P-value for this test is calculated as (using a t-table):

$P-value=2\cdot P(t>1.563)=0.125$

As the P-value (0.125) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the average weekly grocery bill per five-person family in San Antonio is significantly different from the national average.

8 0

3 years ago