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3 years ago

I just need help with the last part. the c) part Describe the end behavior of the function:

Mathematics

2 answers:

Mkey [24]3 years ago

8 0

The end has a more entisified curve than the beginning

IRINA_888 [86]3 years ago

4 0

I'm not sure but I think that you would describe it as positive or steep.

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From 82 books to 138 books<br><br> Find the percent increase. Round to the nearest percent.

FrozenT [24]

Answer:

68%

Step-by-step explanation:

Calculate percentage change

from V1 = 82 to V2 = 138

(V2−V1)|V1|×100

=(138−82)|82|×100

=5682×100

=0.682927×100

=68.2927%change

=68.2927%increase

=68%

3 0

2 years ago

Which of the following best describes the equation below?

WARRIOR [948]

Answer:

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2 years ago

* Example 19<br>Verify that cot 2A + tan 2A = 2 CSC 4A.

dexar [7]

Step-by-step explanation:

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6 0

3 years ago

Help find x. I believe it’s 13

aksik [14]

Answer:

m∠RST = 155°

m∠RSU = 102°

Step-by-step explanation:

m∠RST = m∠RSU + m∠UST

12x - 1 = 9x - 15 + 53

12x - 9x = 1- 15 + 53

3x = 39

x = 39/3

x = 13°

m∠RST = (12x -1)° = (12 * 13 - 1)° = (156 - 1 )° = 155°

m∠RSU = (9x -15)° = (9 * 13 - 15)° = ( 117 - 15)° = 102°

4 0

3 years ago

Assume that a sample is used to estimate a population proportion p. Find the 98% confidence interval for a

Vikki [24]

Answer:

$0.81 - 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.730$

$0.81 + 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.890$

And the confidence interval would be:

$0.730 \leq \p \leq 0.890$

Step-by-step explanation:

Information given:

$n=131$ represent the sample size

$\hat p=0.81$ represent the estimated proportion

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 98% confidence interval the value of $\alpha=1-0.98=0.02$ and $\alpha/2=0.01$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.326$

And replacing into the confidence interval formula we got:

$0.81 - 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.730$

$0.81 + 2.326 \sqrt{\frac{0.81(1-0.81)}{131}}=0.890$

And the confidence interval would be:

$0.730 \leq \p \leq 0.890$

6 0

3 years ago