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4 years ago

Solve the following ten riddle I have 30 ones ,82 thousands ,4 hundred thousand, 60 tens and 100 hundreds what number am I

Mathematics

1 answer:

fredd [130]4 years ago

6 0

Answer:

182,463 is the number

4 hundred thousand
82 thousands
100 hundreds
60 tens
30 ones

Step-by-step explanation:

hope it will help ^_^

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What is "a = 35% times 20"

OLga [1]

The answer is the number 7

4 0

3 years ago

Tue or False? The first distribution shown below had a smaller mean than does the second distribution

Yuliya22 [10]

Answer:

the answer id true

i hope that helps

6 0

4 years ago

Percent of change: from 60 to 64

Contact [7]

Answer:

6.66667% (repeating)

Step-by-step explanation:

Percent change is equal to the change/ original. In this case that is 64-60/60 which is 4/60 which is equal to 0.0666667(repeating). Since they are asking for a percent you have to multiply by 100 and yo will get 6.666667%(repeating)

4 0

4 years ago

H(t)=-10x+6 h(?)=-44

Andreyy89

Answer: h(5)=-44

Step-by-step explanation:

Given

h(t)=-10x+6

h(?)=-44

Substitute the [-44] value to the first equation

-44=-10x+6

Subtract 6 on both sides

-44-6=-10x+6-6

-50=-10x

Divide -10 on both sides

-50/-10=-10x/-10

x=5

Hope this helps!! :)

Please let me know if you have any questions

5 0

3 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 = population proportion of baby girls

(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 One-sample z test for proportions;

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, the test statistics = $\frac{0.54-0.50}{\sqrt{\frac{0.54(1-0.54)}{144} } }$

= 0.96

The value of z test statistics is 0.96.

(f) The P-value of the test statistics is given by;

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

= 1 - 0.8315 = 0.1685

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

(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 we fail to reject our null hypothesis.

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

7 0

3 years ago