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2 years ago

Solve.

Mathematics

2 answers:

olchik [2.2K]2 years ago

5 0

$\huge\text{Hey there!}$

$\huge\boxed{\mathsf{6\times7-3^2\times9+4^3 }}\\\huge\boxed{\mathsf{= 42 - (9)(9) + 4^3}}\\\huge\boxed{\mathsf{= 42 - 81 + 64}}\\\huge\boxed{\mathsf{= -39 + 64}}\\\huge\boxed{\mathsf{= 25}}}}\\\huge\boxed{\mathsf{Therefore, your\ answer\ is: 25}}\huge\checkmark$

$\huge\text{Good luck on your assignment \& enjoy your day!}$

~ $\frak{Amphitrite1040:)}$

Norma-Jean [14]2 years ago

4 0

Answer:

Step-by-step explanation:

6*7-3^2*9+4^3

Exponents first

6*7-9*9+64

Multiply and divide from left to right

42 -9*9+64

42 - 81 +64

Add and subtract from left to right

-39 +64

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An officeSupply company is shipping a case of paper to school. There are 10 reams of paper in the case. If each ream of paper aw

Alenkasestr [34]

------------------------------------
Find 10 reams
------------------------------------
1 ream = 32 ounces
10 reams = 32 x 10
10 reams = 320 ounces

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Convert ounce to pound:
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1 ounce = 0.0625
10 ounces = 0.0625 x 10
10 ounces = 0.625 pounds

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Answer: Each case weighs 0.625 pound.
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3 0

3 years ago

#1) * Find the value of x. Write your answer in simplest form.

Usimov [2.4K]

Answer is 4
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5 0

3 years ago

What is negative five over eight -5/8 b = 10

anastassius [24]

$- \frac{5}{8}\cdot b = 10 \ \ \ |multiply\ both\ sides\ by\ (-\frac{8}{5}) \\\\(-\frac{8}{5})\cdot (- \frac{5}{8})\cdot b =(-\frac{8}{5})\cdot 10\\\\b=(-8)\cdot 2\\\\b=-16$

8 0

3 years ago

In the basket and 12 of them are oranges what percentage of the pieces of fruit in the basket or orange show your work in the bo

notsponge [240]

Answer:

60%

Step-by-step explanation:

There are 20 pieces of fruit in a basket and 12 of them are oranges what percentage of the pieces of fruit in the basket are orange show your work in the box and circle your answer

Total oranges in the basket = 20

Number of oranges = 12

what percentage of the pieces of fruit in the basket are orange

= Number of oranges / Total pieces of fruits × 100

= 12/20 × 100

= 0.6 × 100

= 60%

Percentage of the pieces of fruit in the basket that are oranges = 60%

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

2 years ago