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3 years ago

Please simplify this please it’s due really soon!

Mathematics

1 answer:

iren2701 [21]3 years ago

7 0

Step-by-step explanation:

$\frac{ {13}^{6} \div {13}^{8} \times {13}^{ - 4} }{ {13}^{ - 2} }$

You need to remember the properties of exponents: multiplication adds the exponents, division subtracts the exponents.

If you multiply two powers with the same base, the result will be the base to the sum of the powers.

If you divide two powers with the same base, the result will be the base to the difference of the powers.

In your case you will this:

${13}^{6 - 8 + ( - 4) - ( - 2)}$

That is:

${13}^{ - 4}$

Or:

$\frac{1}{ {13}^{4} }$

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Starting in the 1970s, medical technology allowed babies with very low birth weight (VLBW, less than 1500 grams, about 3.3 pound

cluponka [151]

Answer:

The test statistic value using the VLBW babies as group 1 is z=-2.76±0.01 and the P-value for the test (±0.0001) is 0.0030.

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the proportion of persons with normal birth weight that graduates from high school is significantly greater than the proportion of persons with very low birth weight that graduates from high school.

Then, the null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2> 0$

The significance level is 0.05.

The sample 1 (VLBW group), of size n1=244 has a proportion of p1=0.77049.

$p_1=X_1/n_1=188/244=0.77049$

The sample 2 (control group), of size n2=247 has a proportion of p2=0.79757.

$p_2=X_2/n_2=197/247=0.79757$

The difference between proportions is (p1-p2)=-0.02708.

$p_d=p_1-p_2=0.77049-0.79757=-0.02708$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{188+197}{244+247}=\dfrac{385}{491}=0.988$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.988*0.012}{244}+\dfrac{0.988*0.012}{247}}\\\\\\s_{p1-p2}=\sqrt{0.00005+0.00005}=\sqrt{0.0001}=0.0098$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{-0.02708-0}{0.0098}=\dfrac{-0.02708}{0.0098}=-2.76$

This test is a left-tailed test, so the P-value for this test is calculated as (using a z-table):

$P-value=P(t$

As the P-value (0.0030) is smaller than the significance level (0.05), the effect issignificant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the proportion of persons with normal birth weight that graduates from high school is significantly greater than the proportion of persons with very low birth weight that graduates from high school.

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