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3 years ago

Help what’s the answer

Mathematics

1 answer:

aivan3 [116]3 years ago

7 0

Answer: 2⁴ * 5

<u>Step-by-step explanation:</u>

∧

8 10

∧ ∧

4 2 2 5

∧

2 2

80: 2 x 2 x 2 x 2 x 5

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Oscar has 6 tubular neon lights in order to spell out the word VISION to promote his lighting business.It costs £2 to add a corn

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It is £15 for the words, but I don’t understand the second part

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3 years ago

Simplify: −3xy2(2x2y3 + 7xy2 − 5x3y2)

mestny [16]

collect the like terms

Answer is C

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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

3 years ago

Kevin and his children went into a restaurant and he bought $31.50 worth of hotdogs and drinks. Each hotdog costs $4.50 and each

Musya8 [376]

Answer:

6 hot dogs, 2 drinks

Step-by-step explanation:

let y = number of hot dogs

let x = number of drinks

system:

y = 3x

4.5y + 2.25x = 31.50 (standard form: y = -1/2x + 7)

if you graph the system, lines intersect at (2, 6)

6 0

3 years ago

What is the value of b?<br><br> 0.4(3b+10)=19

Sergio039 [100]

Answer:

12.5

Step-by-step explanation:

Step 1:

0.4 ( 3b + 10 ) = 19 Equation

Step 2:

1.2b + 4 = 19 Multiply

Step 3:

1.2b = 15 Subtract 4 on both sides

Step 4:

15 ÷ 1.2 Divide

Answer:

b = 12.5

Hope This Helps :)

7 0

3 years ago