Evaluate 6-5+5 -(5x3-9) 93 ILL GIVE YOU BRAINLIST AND 15 POINTS

bezimeni [28]

In my opinion i think it’s D.parallelogram

5 0

3 years ago

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.

3 0

3 years ago

How do I find The measure of ML

svlad2 [7]

When two chords intersect inside a circle, the product of the two segments of one chord equals the product of the two segments of the other chord.
MN*NL=KN*LN
10x=4*20=80
x=8
ML=MN+NL=18

4 0

3 years ago

A random sample of 100 automobile owners in thestate of Virginia shows that an automobile is driven onaverage 23,500 kilometers

Anastaziya [24]

Answer:

a) 22497.7 < μ< 24502.3

b) With 99% confidence the possible error will not exceed 1002.3

Step-by-step explanation:

Given that:

Mean (μ) = 23500 kilometers per year

Standard deviation (σ) = 3900 kilometers

Confidence level (c) = 99% = 0.99

number of samples (n) = 100

a) α = 1 - c = 1 - 0.99 = 0.01

$\frac{\alpha }{2} =\frac{0.01}{2}=0.005\\ z_{\frac{\alpha }{2}}=z_{0.005}=2.57$

Using normal distribution table, $z_{0.005$ is the z value of 1 - 0.005 = 0.995 of the area to the right which is 2.57.

The margin of error (e) is given as:

$e= z_{0.005}\frac{\sigma}{\sqrt{n} } = 2.57*\frac{3900}{\sqrt{100} } =1002.3$

The 99% confidence interval = (μ - e, μ + e) = (23500 - 1002.3, 23500 + 1002.3) = (22497.7, 24502.3)

Confidence interval = 22497.7 < μ< 24502.3

b) With 99% confidence the possible error will not exceed 1002.3

5 0

3 years ago

I need help with this it’s saying the “blank” Method would be the best for solving this system of equations

aleksandrvk [35]

Answer:

-12+4y=x..............1

6(-12+4y)+4y=8

-72+24y+4y=8

28y=80

y=3.33

-12+4(3.33)=x

1.32=x

5 0

3 years ago