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zhuklara [117]
3 years ago
8

Find the quotient 2a3+2a2-34a-34/a+1 the number next to the variable are exponets

Mathematics
1 answer:
navik [9.2K]3 years ago
8 0
\frac{2a^3+2a^2-34a-34}{a+1}=  \frac{2a^2(a+1)-34(a+1)}{a+1}= \frac{(a+1)(2a^2-34)}{(a+1)}= 2a^2-34
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The quotient of 27 less than m and twice m
masya89 [10]

Answer:

<u>  m-27 </u>

  2m

Step-by-step explanation:

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Simplify. 1/3(1−1/4)^2 please
Strike441 [17]

Answer:

3/16

Step-by-step explanation:

1/3(1-1/4)^2

1/3(3/4)^2

1/3(9/16)

3/16

4 0
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Through:(-1,-5), slope=5
Travka [436]
What exactly are you looking for?

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3 years ago
• A researcher claims that less than 40% of U.S. cell phone owners use their phone for most of their online browsing. In a rando
antiseptic1488 [7]

Answer:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p > α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

Step-by-step explanation:

Set up hypotheses:

Null hypotheses = H₀: p = 0.40

Alternate hypotheses = H₁: p < 0.40

Determine the level of significance and Z-score:

Given level of significance = 1% = 0.01

Since it is a lower tailed test,

Z-score = -2.33 (lower tailed)

Determine type of test:

Since the alternate hypothesis states that less than 40% of U.S. cell phone owners use their phone for most of their online browsing, therefore we will use a lower tailed test.

Select the test statistic:  

Since the sample size is quite large (n > 30) therefore, we will use Z-distribution.

Set up decision rule:

Since it is a lower tailed test, using a Z statistic at a significance level of 1%

We Reject H₀ if Z < -1.645

We Reject H₀ if p ≤ α

Compute the test statistic:

$ Z =  \frac{\hat{p} - p}{ \sqrt{\frac{p(1-p)}{n} }}  $

$ Z =  \frac{0.31 - 0.40}{ \sqrt{\frac{0.40(1-0.40)}{100} }}  $

$ Z =  \frac{- 0.09}{ 0.048989 }  $

Z = - 1.84

From the z-table, the p-value corresponding to the test statistic -1.84 is

p = 0.03288

Conclusion:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p >  α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

8 0
3 years ago
Can someone pls help me
Alina [70]

Answer:

Sure, what do you need?

4 0
2 years ago
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