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3 years ago

If line tw is parallel to line SX , m ? 1 = 59 , and m ? 6 = 72?, what is m ? 8?

Mathematics

1 answer:

zubka84 [21]3 years ago

8 0

I'm guessing 1 and 4, but I'm not sure.

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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

Consider a data set containing the following values: 70 65 71 78 89 68 50 75 The mean of the preceding values is 70.75. The devi

Leokris [45]

Answer:

If this is the sample data, the sample variance is 125.07 and the sample standard deviation is 11.18

If this is a population data, the population variance is 109.44 and the population standard deviation is 10.46

Step-by-step explanation:

We are given the following data-set:

70, 65, 71, 78, 89, 68, 50, 75

Deviations from the mean

–0.75, –5.75 , 0.25, 7.25, 18.25, –2.75, –20.75, 4.2

Sample size, n = 8

Sample:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

Sum of squares of differences =

0.5625 + 33.0625 + 0.0625 + 52.5625 + 333.0625 + 7.5625 + 430.5625 + 18.0625 = 875.5

$s = \sqrt{\dfrac{875.5}{7}} = 11.18$

$\text{ Sample variance} = s^2 = 125.07$

Population:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

Sum of squares of differences =

0.5625 + 33.0625 + 0.0625 + 52.5625 + 333.0625 + 7.5625 + 430.5625 + 18.0625 = 875.5

$\sigma = \sqrt{\dfrac{875.5}{8}} = 10.46$

$\text{ Population variance} = \sigma^2 = 109.44$

7 0

3 years ago

The voltage V in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance R is slowly increasin

hoa [83]

Answer:

$\frac{dI}{dt}=-0.00016 A/s$

Step-by-step explanation:

We are given that

By ohm's law

$V=IR$

R=389 ohm

I=0.03 A

$\frac{dV}{dt}=-0.06V/s$

$\frac{dR}{dt}=0.07ohm/s$

We have to find rate of change of current I means $\frac{dI}{dt}$

Differentiate the equation w.r.t t

$\frac{dV}{dt}=\frac{dI}{dt}R+I\frac{dR}{dt}$

Substitute the values then we get

$-0.06=\frac{dI}{dt}\times 389+0.03\times 0.07$

$-0.06=389\frac{dI}{dt}+0.0021$

$-0.06-0.0021=389\frac{dI}{dt}$

$-0.0621=389\frac{dI}{dt}$

$\frac{dI}{dt}=\frac{-0.0621}{389}=-0.000160 A/s$

Hence, the current I is changing at the rate=-0.00016A/s

4 0

4 years ago

How to write a book report plz help???

nlexa [21]

Answer:

below

Step-by-step explanation:

In short, a book report is an essay that contains what you read and learned from reading a book. For example, you would write a book report in the following sequence: