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ololo11 [35]
3 years ago
7

Solve using elimination. Brainliest for correct answer!!

Mathematics
1 answer:
dimulka [17.4K]3 years ago
5 0

Answer:

X is 1 and y is 5

Step-by-step explanation:

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Convert 3,200,000 to a scientific notation.
tia_tia [17]
3,200,00 converted to scientific notation is 3.2 x 106
7 0
3 years ago
Read 2 more answers
Write an exponential function in the form y=ab^x that goes through points (0,14) and (3,3024).
Dmitrij [34]

Answer:

y = 14 (6)^{x}

Step-by-step explanation:

For an exponential function of the form y = ab^{x}

Use the given points to find a and b

Using (0, 14 ), then

14 = ab^{0} ( b^{0} = 1 ) , thus

a = 14

y = 14b^{x}

Using (3, 3024 ) , then

3024 = 14 b³ ( divide both sides by 14 )

216 = b³ ( take the cube root of both sides )

b = \sqrt[3]{216} = 6 , thus

y = 14 × 6^{x} ← exponential function

5 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
42 points giv away PLZ ANSWER RN (5/8+3/4) divided by (-2/3 - 5/6)
Triss [41]

Answer:

when you divide you get -11/12

if you have a calculator you could add the 5/8 and 3/4

then add the -2/3 and the -5/6

then divide the first answer by the second answer

8 0
3 years ago
Read 2 more answers
Sandy works at a clothing store. She makes $7 per hour plus earns 10% commission on her sales. She worked 72 hours over the last
MArishka [77]
7(72) + 0.10(2574) = 504 + 257.40 = 761.40 <== this is how much she will earn. Your gonna have to pick the answer choice that is the closest since no answer choices were given.
3 0
3 years ago
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