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2 years ago

ST | PR. Find QR. 35 28 40

Mathematics

1 answer:

Greeley [361]2 years ago

8 0

Answer:

The answer is 47

Step-by-step explanation:

I'm pretty sure this is the answer :)

40 - 28 = 12

35 + 12 = 47

Proof: 35 - 28 = 7

47 - 40 = 7

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If it took 3 people 10 hours to plant a large garden, how long would it take one person? plzz help

ivanzaharov [21]

30 hours. hope this helps :)

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3 years ago

Read 2 more answers

What is 3/8 x 16/7? i don't know. it is for my math homework?

liberstina [14]

Multiply the top and bottom.

3/8 * 16/7 = 48/56

Simplified: 5/7

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3 years ago

(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

7 0

2 years ago

Two number cubes are rolled. What is the probability that the sum of the numbers rolled is either 4 or 8?

sveticcg [70]

There are 6 numbers on each die.

The total number of out comes is 6 x 6 = 36.

To get a sum of 4, the combinations are: 1 & 3, 2 & 2 or 3 & 1.

To get a sum of 8, the combinations are: 1 & 7, 2 & 6, 3 & 5, 4 & 4, 7 & 1, 6 & 2, 5 & 3

The total of those combinations are: 3 + 7 = 10 total combinations of a sum of either 4 or 8.

The probability of either is 10/36, which can be reduced to 5/18.

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3 years ago

How do you do the number line of this question why do we need to flip the inequality

Phantasy [73]

I don't see an inequality, but I can tell you that the sign of the inequality is flipped whenever you multiply or divide by a negative number.

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3 years ago