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morpeh [17]
3 years ago
11

Someone help its multiple choice im so confused

Mathematics
1 answer:
Viktor [21]3 years ago
5 0
A, he had just eaten
B, he was shot in the back
E, the arrow went under his left armpit

This all is correct considering the paragraph i just read!

I wish you luck on the rest of your assessment!

mark this as brainliest if i helped at all!! :)
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wind turbines generate large amounts of electricity from wind energy. the blades on a wind turbine can revolve 22 times per minu
lesya [120]

Answer:

1320 times  

Step-by-step explanation:

            Time = 1 h × (60 min/1 h) = 60 min

Revolutions = 60 min × (22 rev/1 min) = 1320 rev

The blades revolve 1320 times per hour.


5 0
3 years ago
A school puts on a play. The play costs $1,200 in expenses. The students charge $4.00 for tickets. There will be one performance
baherus [9]

Answer:

The domain is the integers from 0 to 500.

Step-by-step explanation:

The idea is that, the auditorium cannot charge for a half seat, and so integers are the only set that can be used.

3 0
3 years ago
Tina ordered a replacement part for her desk. It was shipped in a box that measures 4inches By 4inches by 14 inches. What is the
Sergeeva-Olga [200]

Answer:

  15 inches

Step-by-step explanation:

The space diagonal of a rectangular prism is the root of the sum of the squares of the edge lengths. The longest straight part that will fit in the box will be at most ...

  l = √(4² +4² +14²) = √228 ≈ 15.10 . . . . inches

The greatest length the part could have been is 15 inches.

7 0
3 years ago
Do the information on a bag of fertilizer 5 pounds of fertilizer are needed for every two square feet of lawn how many square fe
gregori [183]

Answer:

25 lb bag of fertilizer can cover <u>10 square feet</u> of lawn.

Step-by-step explanation:

Given:

5 pounds of fertilizer cover = 2 ft^2 of the lawn

We need to find the 25 lb bag cover.

Solution:

Now we know that;

5 pounds of fertilizer = 2  ft^2 of yard.

1 lb of fertilizer = Number of ft^2 lawn covered by 1 lb fertilizer.

Using Unitary method we get;

Number of ft^2 lawn covered by 1 lb fertilizer = \frac{2}{5}\ ft^2

Now we can say that;

1 lb of fertilizer =  \frac{2}{5}\ ft^2 of the yard

25 lb of fertilizer = Number of ft^2 lawn covered by 25 lb fertilizer.

Again by using Unitary method we get;

Number of ft^2 lawn covered by 25 lb fertilizer. = \frac{2}{5}\times 25 = 10 \ ft^2

Hence 25 lb bag of fertilizer can cover <u>10 square feet</u> of lawn.

7 0
3 years ago
In a sample of nequals16 lichen​ specimens, the researchers found the mean and standard deviation of the amount of the radioacti
GuDViN [60]

Answer:

(a) The confidence level desired by the researchers is 95%.

(b) The sampling error is 0.002 microcurie per millilitre.

(c) The sample size necessary to obtain the desired estimate is 25.

Step-by-step explanation:

The mean and standard deviation of the amount of the radioactive​ element, cesium-137 present in a sample of <em>n</em> = 16 lichen specimen are:

\bar x=0.009\\s=0.005

Now it is provided that the researchers want to increase the sample size in order to estimate the mean μ to within 0.002 microcurie per millilitre of its true​ value, using a​ 95% confidence interval.

The (1 - <em>α</em>)% confidence interval for population mean (μ) is:

CI=\bar x\pm z_{\alpha/2}\times \frac{s}{\sqrt{n}}

(a)

The confidence level is the probability that a particular value of the parameter under study falls within a specific interval of values.

In this case the researches wants to estimate the mean using the 95% confidence interval.

Thus, the confidence level desired by the researchers is 95%.

(b)

In case of statistical analysis, during the computation of a certain statistic, to estimate the value of the parameter under study, certain error occurs which are known as the sampling error.

In case of the estimate of parameter using a confidence interval the sampling error is known as the margin of error.

In this case the margin of error is 0.002 microcurie per millilitre.

(c)

The margin of error is computed using the formula:

MOE=z_{\alpha/2}\times \frac{s}{\sqrt{n}}

The critical value of <em>z</em> for 95% confidence level is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

*Use a <em>z</em>-table.

MOE=z_{\alpha/2}\times \frac{s}{\sqrt{n}}

 0.002=1.96\times \frac{0.005}{\sqrt{n}}

       n=[\frac{1.96\times 0.005}{0.002}]^{2}

          =(4.9)^{2}\\=24.01\\\approx 25

Thus, the sample size necessary to obtain the desired estimate is 25.

6 0
4 years ago
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