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Karo-lina-s [1.5K]

3 years ago

11

At 3pm corettas shadow is 1.03 meters long. Her height is 1.65 meters at the same time, a trees shadow is 5.57 meters long. How

tall is the tree

1 answer:

Komok [63]3 years ago

3 0

3.477 meters long...................

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5 liters of water weighs 11 pounds. A bucket contains 143 pounds of water. How many liters of water are in the bucket

slavikrds [6]

Answer:

65 liters

Step-by-step explanation:

11/5=2.2

143/2.2=65

4 0

3 years ago

Daryl has a total of 160 National Football League (NFL) trading cards from both the American Football Conference (AFC) and the N

Svet_ta [14]

Answer:

B

Step-by-step explanation:

Since A represents the number of AFC cars and N represents the number of NFC cards, and that Daryl has 160 total cards, then A+N = 160, the total number of cards. Furthermore, the number of AFC cards Daryl has is 8 more than 3 times the number of NFC cards, so A would be equal to 8 + 3N, or A = 3N + 8. The only system of equations with both of these correct equations is B.

4 0

3 years ago

Simplify (-6)^5 how do you simplify the the exponent expression

Papessa [141]

X^m means x times itself m times

(-6)^5 means -6 time itself 5 times
simplify it says
remember that
(mn)^x=(m^x)(n^x)
undistribute the -1
(-6)^5=(-1)^5 times (6^5)
(-1)^5=-1 since it is odd power
can be simplified to
-6^5
(note, -6^5 vs (-6)^5 PEMDAS)

5 0

3 years ago

Read 2 more answers

(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

The amount of revenue for a business can be modeled by the function

GarryVolchara [31]

God loves all of you, remember that oK. Thts really all u need in life <3 7 if u know answer in comments - 8400(3)2t= H is

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2 years ago