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3 years ago

I need to knw ASAP PLZ HELP ME OUT.

Mathematics

2 answers:

Anastaziya [24]3 years ago

6 0

Answer:

Step-by-step explanation:

this makes no sense you can subtrct anything by nothig to make 1 this is rigged

Pie3 years ago

4 0

I think it is 23.27-5.3 equals 17.97

I checked and this works

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How many different combinations of shirts and ties are possible if you have 4 shirts and 5 ties?

Fittoniya [83]

20 different combinations.

8 0

3 years ago

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Find the unit price

Mrac [35]

Answer:

Price per ounce is $0.062

Step-by-step explanation:

Here, we want to find the unit price in which this vase is price per ounce

Firstly, we need the total ounce value

1 pound = 16 ounce

so two pounds will be 2 * 16 = 32 ounces

So the total ounce value is 32 ounces + 8 ounce = 40 ounces

So 40 ounces = $2.48

price per ounce will be $2.48/40 = $0.062 per ounce

3 0

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

3 years ago

Work out the value of angle x

Alexus [3.1K]

well, it's isosceles so use base angles theorem

the top angle is also x

90 + 2x = 180

subtract 90 from both sides

2x = 90

divide both sides by 2

x = 45 degrees

8 0

3 years ago

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When solving a system of two linear equations

soldier1979 [14.2K]

Answer:

When both equations have the same slope, but not the same y-intercept, they'll be parallel to each other and no intersections means no solutions. When both equations have different slopes than regardless of the y-intercept they'll intersect for certain, therefore it has exactly one solution.

Step-by-step explanation:

Got this from google hope it helps

7 0

3 years ago