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3 years ago

A tiger can eat up to 55 pounds a meat a day about hiw many days would it take to eat the follwing pray

Mathematics

1 answer:

Sholpan [36]3 years ago

4 0

May you give answer choices please.

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Help plz it’s 50 points help

abruzzese [7]

Answer:

Step-by-step explanation:

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3 years ago

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Which is a possible expression for the series <br> -6-12-24-48-96-192

iren [92.7K]

Answer:

Math is confusing

Step-by-step explanation:

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3 years ago

Refer to the figure to complete the proportion A/C = Y/?

salantis [7]

Hello,
Please, see the attached file.
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3 years ago

You want to obtain a sample to estimate how much parents spend on their kids birthday parties. Based on previous study, you beli

zhuklara [117]

Answer:

We need to sample at least 1069 parents.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.05 = 0.95$ , so $z = 1.645$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How many parents do you have to sample?

We need to sample at least n parents.

n is found when $M = 4, \sigma = 79.5$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$4 = 1.645*\frac{79.5}{\sqrt{n}}$

$4\sqrt{n} = 1.645*79.5$

$\sqrt{n} = \frac{1.645*79.5}{4}$

$(\sqrt{n})^{2} = (\frac{1.645*79.5}{4})^{2}$

$n = 1068.92$

Rounding up

We need to sample at least 1069 parents.

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3 years ago

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IgorC [24]

Answer:

Answer is -8.........bro

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3 years ago