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Sergio [31]
3 years ago
13

Please help me with measurements

Mathematics
1 answer:
levacccp [35]3 years ago
4 0
I dont know the answer but you may find this helpful:
-1 Quart = 1.13 British litres
-Fluid ounce = 0.028 Brtish litres
-1 gallon = 4.546 British litres
-1 Cup is 0.237 British litres

Now all the mesurments are the same i hope this helps a bit.
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Pls help I will mark brainliest if it’s right
vladimir2022 [97]

Answer:

C

Step-by-step explanation:

3 0
3 years ago
If Uber charges $12 for a ride plus $2 per mile, write an equation for the cost (where x is the miles)
Luba_88 [7]

Answer:

12 + 2x = y

Step-by-step explanation:

y would be the total cost

7 0
3 years ago
Read 2 more answers
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
Answer all parts in the question below. Be sure to show all your work. Remember NO WORK, NO CREDIT EVEN IF THE ANSWER IS CORRECT
liraira [26]

I'm sorry I need the points lol anyways I hope you got the answer to it

3 0
3 years ago
I need help with this pls I out 30 points<br>​
KonstantinChe [14]

Answer:

Step-by-step explanation:

Age:            Frequency:

50-54          2

55-59          5

60-64          3

65-69          3

6 0
3 years ago
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