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3 years ago

Find the sum of the three expressions and choose the correct answer.

Mathematics

2 answers:

riadik2000 [5.3K]3 years ago

7 0

Answer:

Sum of -8a³b+3a²b²-5,2+5a²b² and 3a³b+7-9a²b² is:

-5a³b-a²b²+4

Step-by-step explanation:

We have to find the sum of:

-8a³b+3a²b²-5,2+5a²b² and 3a³b+7-9a²b²

-8a³b+3a²b²-5+2+5a²b²+3a³b+7-9a²b²

= -5a³b-a²b²+4

Hence, sum of -8a³b+3a²b²-5,2+5a²b² and 3a³b+7-9a²b² is:

-5a³b-a²b²+4

cupoosta [38]3 years ago

6 0

<span>-8a^3 b + 3a^2 b^2 - 5 + 2 + 5a^2 b^2 + 3a^3 b + 7 - 9a^2 b^2
= </span>-5a^3 b - a^2 b^2 + 4

answer is

<span>C.) -5a3b - a2b2 + 4</span>

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And the 90% confidence interval would be given (0.131;0.169).

Step-by-step explanation:

We have the following info given:

$n= 955$ represent the sampel size slected

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The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

$\hat p =\frac{955-812}{955}= 0.150$

The confidence interval for the proportion would be given by this formula

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

For the 90% confidence interval the significance is $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution and we got.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131$

$0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169$

And the 90% confidence interval would be given (0.131;0.169).

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