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tigry1 [53]
4 years ago
11

There are a total of 64 students in a drama club and a yearbook club. The drama club has 14 more students than the yearbook club

.a. Write a system of linear equations that represents this situation. Let x represent the number of students in the drama club and y represent the number of students in the yearbook club.
Mathematics
1 answer:
Scorpion4ik [409]4 years ago
7 0
L=x which repents students
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Question 2 is B because 55 in fraction is 55/100 simplified by 11/20
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3 years ago
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pochemuha

Answer:5

Step-by-step explanation:

just tryna get points but it goes down by 3 eachtime

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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
I am really stuck on this problem​
oksian1 [2.3K]

Answer:

1. ∠1 = 60

2. ∠2 = 40

3. ∠3 = 80

4. ∠4 = 80

5. ∠5 = 60

6. ∠6 = 120

7. ∠7 = 100

8. ∠8 = 60

9. ∠9 = 120

10. ∠10 = 100

Step-by-step explanation:

8 0
3 years ago
What is the answerrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr
Alecsey [184]

Answer:

$1559.40

Step-by-step explanation:

1. Find the perimeter of the garden

14\frac{2}{3} +  14\frac{2}{3} + 10\frac{1}{3} + 10\frac{1}{3} = 50 yd

2. Convert to feet

1 yd = 3ft

50 x 3 = 150

50 yd = 150 ft

3. Divide by 2.5 to find out how many sections are needed

150/2.5 = 60 sections

4.  Multiply by 25.99

60 x 25.99 = 1559.4

It will cost $1559.40

3 0
3 years ago
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