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Svetach [21]
3 years ago
12

Which option summarizes what gave Chandra Gupta I power

Mathematics
1 answer:
mars1129 [50]3 years ago
6 0

Answer:

Chandra Gupta I, King of India reigned from 320 to about 330 CE and founder of the Gupta empire. He was the grandson of Sri Gupta, the first known leader of the Gupta line. Chandra Gupta I, whose first life is unknown, became the local chief of the kingdom of Magadha parts of the modern state of Bihar.

Step-by-step explanation:

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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
Y - 18 = 2/7 (x + 6)
r-ruslan [8.4K]

Answer:

y=2/7x+138/7

Step-by-step explanation:

hope this helps!

7 0
3 years ago
(Giving 80 points, Please do not answer if you don't know. The second and third answer are not 4 btw)
Jobisdone [24]

Answer:

1. 4

2. 10

3. 14

Step-by-step explanation:

A = -6

B = 0

C = 4

D = 8

1. BC = |0 - 4| = 4

2. AC = |-6 - 4| = |-10| = 10

3. AD = |-6 - 8| = |-14| = 14

3 0
2 years ago
Read 2 more answers
To multiply a monomial by a polynomial, use the (Associative or Distributive) Property. Using this property, the product of (2k^
JulijaS [17]
I'd do this in reverse order.  Mult. each of the 3 terms inside parentheses by 4:

8k^2 - 28k + 12


6 0
3 years ago
Please I need help! I'll give brainiest please! And explain how you got your answer please!
Mrac [35]
The answer is 27 or D
3 0
3 years ago
Read 2 more answers
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