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3 years ago

Help me please!! I’ve tried many different things and nothing will work

Mathematics

1 answer:

Arte-miy333 [17]3 years ago

7 0

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Right answer gets brainlist

Alika [10]

90. you can just do 5x12 which is 60. 60x3 is 180 then 180 divided by 2 because a triangular prism is just half of a cube

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2 years ago

Read 2 more answers

If P(E)=0.55, P(E or F)=0.65, and P(E and F)=0.20, find P(F).

sleet_krkn [62]

Answer:

Step-by-step explanation:

let x= P(f)

.65= .55+x-.2

P(F)=.3

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3 years ago

How much will you pay for 5 pounds of shrimp if the cost is 2 pounds for $10.99? show your work.

Mkey [24]

Cost for 2 pounds of shrimp = 10.99
Cost for 5 pounds of shrimp
(10.99 / 2) * 5
$ 27.475

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3 years ago

Production of passenger cars in Japan increased from 3.94 million in 1999 to 6.74 million in 2009. What is the geometric mean an

vovangra [49]

Let's assume initial population is in 1999

so, production of passenger cars in Japan is 3.94 million in 1999

so, P=3.94 million

and the production of passenger cars in Japan is 6.74 million in 2009

so, A=6.74 million in t=2009-1999=10 years

now, we can use formula

$A=P(1+r)^t$

here , r is interest rate

so, we can plug values

$6.74=3.94(1+r)^{10}$

now, we can solve for

$\left(1+r\right)^{10}=\frac{337}{197}$

$r=0.05516$

so,

the geometric mean annual percent increase is 5.516%............Answer

7 0

3 years ago

Suppose that the epidemiologist wants to re-estimate the population proportion and wishes for her 95% confidence interval to hav

Svetllana [295]

Answer:

She needs a sample of at least 385.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

How large a sample should she take to achieve this?

She needs a sample of size at least n.

n is found when M = 0.04.

We do not know the true population proportion, so we use $\pi = 0.5$ , which is the case for which we are going to need the largest sample size.

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.05 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.05\sqrt{n} = 1.96*0.5$

Dividing both sides by 0.05

$\sqrt{n} = 19.6$

$(\sqrt{n})^{2} = 19.6^{2}$

$n = 384.2$

Rounding up

She needs a sample of at least 385.

3 0

3 years ago