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2 years ago

Find the indicated limit, if it exists. See file below.

Mathematics

1 answer:

White raven [17]2 years ago

6 0

Answer:

Step-by-step explanation:

The limit exists because it's continuous since they equal each other

when x approaches zero.

Substitute both expressions with x=0 and both of them will equal 2, so the limit should be 2.

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Read 2 more answers

Suppose a large shipment of microwave ovens contained 12% defectives. If a sample of size 474 is selected, what is the probabili

melisa1 [442]

Answer:

$P(\hat p>0.14)$

And using the z score given by:

$z = \frac{\hat p -\mu_p}{\sigma_p}$

Where:

$\mu_{\hat p} = 0.12$

$\sigma_{\hat p}= \sqrt{\frac{0.12*(1-0.12)}{474}}= 0.0149$

If we find the z score for $\hat p =0.14$ we got:

$z = \frac{0.14-0.12}{0.0149}= 1.340$

So we want to find this probability:

$P(z>1.340)$

And using the complement rule and the normal standard distribution and excel we got:

$P(Z>1.340) = 1-P(Z$

Step-by-step explanation:

For this case we have the proportion of interest given $p =0.12$ . And we have a sample size selected n = 474

The distribution of $\hat p$ is given by:

$\hat p \sim N (p , \sqrt{\frac{p(1-p)}{n}})$

We want to find this probability:

$P(\hat p>0.14)$

And using the z score given by:

$z = \frac{\hat p -\mu_p}{\sigma_p}$

Where:

$\mu_{\hat p} = 0.12$

$\sigma_{\hat p}= \sqrt{\frac{0.12*(1-0.12)}{474}}= 0.0149$

If we find the z score for $\hat p =0.14$ we got:

$z = \frac{0.14-0.12}{0.0149}= 1.340$

So we want to find this probability:

$P(z>1.340)$

And using the complement rule and the normal standard distribution and excel we got:

$P(Z>1.340) = 1-P(Z$

4 0

3 years ago

Please help I have no clue how to do this

Alinara [238K]

Step-by-step explanation:

It's a question of trigonometry.

You need to remember that tangent (tan) stands for Perpendicular / Base.

So, tan P = QR / PR

By Pythagoras Theorem,

34² = 30² + x²

x² = 34² - 30²

x² = 1156 - 900

x² = 256

x = 16

Now, placing the values,

tan P = 30 / 16

tan P = 1.875

HOPE IT HELPS ^_^

5 0

2 years ago