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2 years ago

In 2015, there were approximately 33,400 firearm fatalities (homicide and suicides) Express this quantity in deaths per hour

1 answer:

4 0

8765.82 are how many hours there are in a year take that divided by 33400 and you get 3.81025392 or rounded up it would be 4 deaths per hour or just 3.8 per hour

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5. Find the mZABC if<br> ZABD = 35°, ZDBC = 55º.

Reika [66]

Answer:

$m\angle ABC=90^o$

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

$m\angle ABC=m\angle ABD+m\angle DBC$ ----> by angle addition postulate

substitute the given values

$m\angle ABC=35^o+55^o=90^o$

5 0

3 years ago

You deposit $800 in an account that pays 2.65% annual interest compounded continuously. What is the balance after 12 years? Use

balu736 [363]

Answer:Let P = initial investment

r = annual interest rate (decimal form)

t = number of years

A(t) = amount after t years

Then, A(t) = Pert

A(12.5) = 800e(0.0265)(12.5)

= 800e0.33125

= $1114.17

Step-by-step explanation:

4 0

2 years ago

Michael

KengaRu [80]

Answer:

I believe the answer should be 90%

4 0

3 years ago

23. A certain type of gasoline is supposed to have a mean octane rating of at least 90. Suppose measurements are taken on of 5 r

Nostrana [21]

Answer:

a) $p_v =P(Z$

b) Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

c) $P(\bar X >90)=1-P(Z$

$1-P(Z$

d) For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

Step-by-step explanation:

Previous concepts and data given

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".

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

$\bar X=88.96$ represent the sample mean

$s=1.011$ represent the sample standard deviation

n=5 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 90, the system of hypothesis would be:

Null hypothesis: $\mu \geq 90$

Alternative hypothesis: $\mu < 90$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{88.96-90}{\frac{0.8}{\sqrt{5}}}=-2.907$

Part a

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=5-1 = 4$

Then since is a left tailed test the p value would be:

$p_v =P(Z$

Part b

Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

Part c

We want this probability:

$P(\bar X$

The best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

If we apply this formula to our probability we got this:

$P(\bar X >90)=1-P(Z$

$1-P(Z$

Part d

For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

7 0

3 years ago

What radius of a circle is required to inscribe a regular hexagon with an area of 166.28 in2 and an apothem of 6.928 in?

Lunna [17]

If you want to inscribe a polygon inside a circle, you have a formula that doesn't have to use the apothem. The formula is:

A = (nr²/2)sin(360/n)

Since the polygon is a hexagon, it has 6 sides. Thus, n = 6. Knowing the area, we can determine the radius of the circle, r.

166.28 = (6r²/2)sin(360/6)
r = 8 inches

Thus, the radius of the circle is 8 inches.

7 0

3 years ago

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