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2 years ago

14

You are playing a video game. For each diamonds you collect, you earn 50 points. You need to earn 2000 points or more to continu

e to the next level. The variable d represents the number of diamonds your collect

1 answer:

Brrunno [24]2 years ago

4 0

40 diamonds x 50 point (each diamond) = 2000 points

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A customer at Lost Pastores placed an order for 3 burritos which was priced at x amount of dollars each. He also ordered a plate

timofeeve [1]

Answer:

3x +12.95 = 32.46

Step-by-step explanation:

3 burritos plus x + 12.95 = $32.46

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2 years ago

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Simply 3 26/27 + 2 3/4 + 15 1/9

cupoosta [38]

1) 3+2+15=20
2) 3/4= 27/36 1/9=4/36
3) 27/36+4/36=31/36
4) 26/27=936/972 31/36= 637/972
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What is a factor of -270 that adds to be -39

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For this one your answer is -309

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3 years ago

Last Sunday, the average temperature was 8%, percent higher than the average temperature two Sundays ago. The average temperatur

MArishka [77]

Answer: 1.08T and (1+8/100)T

Step-by-step explanation:

Given: The average temperature two Sundays ago was T degrees Celsius.

Since, last Sunday, the average temperature was 8% higher than the average temperature two Sundays ago.

⇒ The average temperature last Sunday $=8\%\ of\ T+T$

Since 8%=0.08

Therefore, The average temperature last Sunday $=\frac{8}{100}\timesT+T$

⇒ The average temperature last Sunday $=T(\frac{8}{100}+1)$

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⇒ The average temperature last Sunday $=T(1.08)$

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2 years ago

In a sample of 42 water specimens taken from a construction site, 26 contained detectable levels of lead. Construct 90% confiden

hram777 [196]

Answer:

(0.4958, 0.7422)

Step-by-step explanation:

Let p be the true proportion of water specimens that contain detectable levels of lead. The point estimate for p is $\hat{p}=26/42=0.6190$ . The estimated standard deviation is given by $\sqrt{\hat{p}(1-\hat{p})/n}=\sqrt{(0.6190)(1-0.6190)/42}=0.0749$ . Because we have a large sample, the 90% confidence interval for p is given by $0.6190\pm z_{0.05}0.0749$ where $z_{0.05}=1.6448$ is the value that satisfies that above this and under the standard normal density there is an area of 0.05. So, the confidence interval is $0.6190\pm (1.6448)(0.0749)$ , i.e., (0.4958, 0.7422).

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3 years ago