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Ad libitum [116K]
2 years ago
15

2. Order the numbers from smallest to largest. *

Mathematics
1 answer:
Ostrovityanka [42]2 years ago
4 0

Answer:

1. 0.7

2. 0.75

3. 23

Step-by-step explanation:

You didn't provide a fourth number.

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PLZ HEEEEEEEELPPPPPPPP!!!!!!!!!
Elena L [17]
Y= ∛(-x) - 3
for x = -8 → y = ∛-(-8) - 3 ↔ y = 2-3 ↔y = -1
for x= 8 → y= ∛(-8) - 3 ↔ y = -2 -3 ↔ y = -5
Then  - 5≤ y≤ - 1
hence the range of y is {y|-5≤y≤-1}
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
2 years ago
point X is located at (2,-6), and point Z is located at (0,5). find the value for the point Y that is located 1/5 the distance f
HACTEHA [7]
Let point is(x,y)
x=(2-0)/5=2/5
y=(-6-5)/5=-11/5
(2/5,-11/5)
8 0
3 years ago
HELP ME PLZ!!!!! Plz!!!
Charra [1.4K]

Answer:

hi the right answer is option C.112

here's why:

\frac{14}{50}  \times 400 = 112

4 0
2 years ago
Jamie had $37 in his bank account on Sunday the table shows his account activity for the next 4 days what was the balance and Ja
Ilya [14]
What is the table showing?
5 0
3 years ago
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