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nikdorinn [45]
4 years ago
7

Neo could choose from 3 different pants, 5 different shirts, and 4 sweaters. How many different combinations of 1 pant, 1 shirt

and 1 sweater could he choose?
Mathematics
1 answer:
Sidana [21]4 years ago
7 0
To answer you do 5 times 4 times 3 to get 60
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Tasha spends 25 minutes reading on Wednesday night.She spends 17 more minutes reading on Thursday than she did on Wednesday. Wri
Harman [31]

Answer:

25+17=42

Step-by-step explanation:

so you add 25 and 17 and get 42, and thats how many minutes she read

how do you not know this in high school

7 0
3 years ago
Read 2 more answers
Find the value of y.
grandymaker [24]

y^2 = 9(3+9)\\\\y^2 = 3^2\cdot3\cdot2^2\\\\y = 6\sqrt{3}

5 0
4 years ago
A 0.1 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender​ selecti
Andrews [41]

Answer:

(a) Null Hypothesis, H_0 : p = 0.50

    Alternate Hypothesis, H_A : p > 0.50

(b) The value of level of significance (α) given in the question is 0.10.

(c) The sampling distribution of the sample statistic is Normal distribution.

(d) This test is right-tailed.

(e) The value of z test statistics is 0.96.

(f) The P-value is 0.1685.

(g) At 0.10 significance level the z table gives critical value of 1.282 for right-tailed test.

(h) We conclude that the proportion of baby girls is equal to 0.50.

Step-by-step explanation:

We are given that a 0.1 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender​ selection, the proportion of baby girls is greater than 0.5.

Assume that sample data consists of 78 girls in 144 ​births.

Let p = <u><em>population proportion of baby girls</em></u>

(a) So, Null Hypothesis, H_0 : p = 0.50     {means that the proportion of baby girls is equal to 0.50}

Alternate Hypothesis, H_A : p > 0.50     {means that the proportion of baby girls is greater than 0.50}

(b) The value of level of significance (α) given in the question is 0.10.

(c) The sampling distribution of the sample statistic is Normal distribution.

(d) This test is right-tailed as in the alternative hypothesis we are concerned for proportion of baby girls that is greater than 0.50.

(e) The test statistics that would be used here <u>One-sample z test for proportions</u>;

                             T.S. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of baby girls =  \frac{78}{144} = 0.54

            n = sample of births = 144

So, <u><em>the test statistics</em></u>  =  \frac{0.54-0.50}{\sqrt{\frac{0.54(1-0.54)}{144} } }  

                                       =  0.96

The value of z test statistics is 0.96.

(f) <u>The P-value of the test statistics is given by;</u>

            P-value = P(Z > 0.96) = 1 - P(Z < 0.96)

                          = 1 - 0.8315 = 0.1685

<u></u>

(g) <u>Now, at 0.10 significance level the z table gives critical value of 1.282 for right-tailed test.</u>

(h) Since our test statistic is less than the critical value of z as 0.96 < 1.282, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region (which was to the right of value of 1.282) due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that the proportion of baby girls is equal to 0.50.

7 0
3 years ago
Find the equation of the median from B, in ABC whose vertices are A(1, 5), B(5, 3), and
VMariaS [17]

Answer:

Step-by-step explanation:

Midpoint of AC = M(-1, 1.5)

Slope of BM = (1.5 - 3)/(-1 - 5) = ¼

Point-slope equation for line of slope ¼ that passes through B(5,3):

y-3 = ¼(x-5)

7 0
3 years ago
the student rate for ymca on broadway is 64.00 per month in order to join there is a refundable $15 registration fee what is the
Snowcat [4.5K]

Answer:

568

Step-by-step explanation:

because we multiply the 64 with the number that will be in the number

7 0
3 years ago
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