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Sophie [7]
3 years ago
5

What is the interquartile range of the following data set? 46, 9, 48, 96, 61, 84, 29, 1, 82, 5, 34

Mathematics
1 answer:
Xelga [282]3 years ago
6 0
First we put the numbers in order

1,5,[9],29,34,(46),48,61,[82],84,96

Q1 = 9
Q2 = 46
Q3 = 82

interquartile range (IQR) = Q3 - Q1 = 82 - 9 = 73 <== that is the correct answer if u are not removing the outliers
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Joshua currently does a total of 8 pushups each day. He plans to increase the number of pushups he does each day by 2 pushups un
Maslowich

Answer:

x=12

Step-by-step explanation:

Let y equal number of pushups a day. When x = 1, y = 8. When x = 2, y = 10. 10 is a factor of 30. Multiply y by 3. However, since x increases by 1 when y increases by 2, multiply x by 6 (2*3). That equals 12. Therefore, when x = 12, y = 30.

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7 0
4 years ago
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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.

3 0
3 years ago
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Answer:

(a) The ones that are equivalent to the given fraction are: \frac{2}{-9} and -\frac{2}{9}

(b) The one that is equivalent to the given fraction is: \frac{-8}{-5}

3 0
2 years ago
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