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MA_775_DIABLO [31]
3 years ago
15

Identify the x and y coordinates of the ordered pair (-5, 9)

Mathematics
1 answer:
valentina_108 [34]3 years ago
3 0

Answer:

x= -5 and y= 9

Step-by-step explanation:

In the writing of coordinates, the coordinate is always the first number in the set and the y coordinate is always the second number.

that is (x, y)= (-5 , 9)

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Which expression is equivalent and why?
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Answer:

  C.  1.15x

Step-by-step explanation:

You are given the expression for the cost:

  x + 0.15x

Combining like terms, this becomes ...

  1.15x . . . . . matches choice C

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Previously, an organization reported that teenagers spent 24.5 hours per week, on average, on the phone. The organization thinks
Lena [83]

Answer:

We need to develop a one-tail t-student test ( test to the right )

We reject H₀  we find evidence that student spent more than 24,5 hours on the phone

Step-by-step explanation:

Sample size  n = 15     n < 30

And we were asked if the mean is higher than, therefore is a one-tail t-student test ( test to the right )

Population mean   μ₀  = 24,5

Sample mean   μ  =  25,7

Sample standard deviation s = 2

Hypothesis Test:

Null Hypothesis      H₀                             μ  =  μ₀

Alternative Hypothesis     Hₐ                  μ  >  μ₀

t (c) =  ?

We will define CI = 95 %  then   α = 5 %   α = 0,05    α/2 =  0,025

n = 15     then degree of freedom    df = 14

From t-student table  we get:  t(c) = 2,1448

And  t(s)

t(s) = ( μ  -  μ₀  ) / s/√n

t(s) = (25,7 - 24,5) /2/√15

t(s) = 2,3237

Now we compare   t(c)   and  t(s)

t(c)  =  2,1448         t(s)  = 2,3237

t(s) > t(c)

Then we are in the rejection region we reject H₀   we have evidence at 95% of CI that students spend more than 24,5 hours per week on the phone

8 0
2 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
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