1.

3. QR find the arc length 02.83 021.99 O 12.57 0 34.56

jeka94

Your answer should be 12.57

8 0

3 years ago

Write three million five hundred twenty six thousand two hundred forty five in standard form

Temka [501]

3,526,245 is the answer.

3 0

3 years ago

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The radius of a circle with area A can be approximated using the formula r = the square root of A/3 . Estimate the radius of a w

Arturiano [62]

Answer: 12

Step-by-step explanation:

Given:

A = 452

r = √A/3

Step 1: Substitute 452sqf for A

r = √452/3

Step 2: Solve

452 ÷ 3 = 150.66∞

√150.66∞ = 12.27∞

12.27∞ rounded to the nearest integer equal 12

3 0

3 years ago

If you pull eight coins from your pocket, and none of them are pennies, what is the most money that you could have? The least?

kramer

If you're based in the UK, you could say up to £800 as we have £100 coins here.

Lowest amount? Besides our penny our least valuable coin is the 2 penny coin. So barring pennies the least you could pull out of your pocket if you're based in the UK is 16p.

5 0

3 years ago

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A local bottler in Hawaii wishes to ensure that an average of 22 ounces of passion fruit juice is used to fill each bottle. In o

Mandarinka [93]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ = 22 ounces

Alternate Hypothesis, $H_A$ : $\mu\neq$ 22 ounces

(b) The value of the test statistic is -2.687.

(c) The critical values are -1.96 and 1.96.

(d) We conclude that Reject $H_0$ since the value of the test statistic is less than the negative critical value.

Step-by-step explanation:

We are given that a local bottler in Hawaii wishes to ensure that an average of 22 ounces of passion fruit juice is used to fill each bottle.

He takes a random sample of 65 bottles. The mean weight of the passion fruit juice in the sample is 21.54 ounces. Assume that the population standard deviation is 1.38 ounce.

Let $\mu$ = average ounces of passion fruit juice used to fill each bottle.

(a) So, Null Hypothesis, $H_0$ : $\mu$ = 22 ounces {means that the average of 22 ounces of passion fruit juice is used to fill each bottle}

Alternate Hypothesis, $H_A$ : $\mu\neq$ 22 ounces {means that the average different from 22 ounces of passion fruit juice is used to fill each bottle}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean weight of the passion fruit juice = 21.54 ounces

$\sigma$ = population standard deviation = 1.38 ounce

n = sample of bottles = 65

So, test statistics = $\frac{21.54-22}{\frac{1.38}{\sqrt{65} } }$

= -2.687

(b) The value of z test statistics is -2.687.

(c) Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that Reject $H_0$ since the value of the test statistic is less than the negative critical value which means that the average different from 22 ounces of passion fruit juice is used to fill each bottle.

6 0

3 years ago