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3 years ago

10.5 X 3.65=? Simplify

Mathematics

2 answers:

34kurt3 years ago

8 0

Answer:

38.325

i did the eqation

MrRissso [65]3 years ago

7 0

Answer:

38.3

Step-by-step explanation:

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Sadie and Connor both play soccer. Connor scored 2 times as many goals as Sadie. Together they scored 9 goals. Could Sadie have

astraxan [27]

Answer: No, because 4(2) + 4 ≠ 9

Step-by-step explanation:

No, because 4(2) + 4 ≠ 9

7 0

3 years ago

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Can someone help me do this darn work plz

PSYCHO15rus [73]

Answer:

13. 17/20

14. 11/100

16.) 4 3/10

17.) 7 3/4

19.) -1 3/5-

20.) 3/8

22.) -5 3/5

23.) 1 3/25

26.) 85.33%

5 0

3 years ago

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Tom bought 3 gallons of grey paint and 2 gallons of red paint. Grey paint is $24.59 per gallon and red paint is $37.99 per gallo

tigry1 [53]

Answer:

The red will cost him more by $2.21

Step-by-step explanation:

First you multiply the amount of gallons of grey paint and the cost and you get 24.59 x 3= 73.77

The do the same for red

37.99 x 2= 75.98

the red is going to cost him more by 75.98-73.77= $2.21

6 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

In a game, four cards are labeled N, S, E, and W. Two tiles are numbered 1 and 2. Two

nikitadnepr [17]

Answer:1/16

Step-by-step explanation:

1/4x1/2x1/2

8 0

4 years ago

Read 2 more answers