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dangina [55]
4 years ago
7

Two thirds a number d to the third power

Mathematics
1 answer:
In-s [12.5K]4 years ago
7 0
((2d)/3)^3=(2d)/3*(2d)/3*(2d)/3=(8d^3)/27

Sorry if this is confusing
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Use properties of rational numbers to multiply the following. -45 x (-85)
Marizza181 [45]
Multiplying -45 and -85
\frac{ - 45}{1}   \times  \frac{ - 85}{1}  =  \frac{ - 45 \times  ( - 85)}{1 \times 1}  =  \frac{3825}{1}
So 3825 is the final answer.
4 0
3 years ago
he temperature dropped 12°F in 8 hours.If the final temperature was -7°F, what was the starting temperature?
anygoal [31]

Answer:

5°F

Step-by-step explanation:

The starting temperature is 5.

You get this answer when you add 12 to -7.

If the question is saying the temperature dropped 12°F every 8 hours then the answer would be 89°F because you multiply 12 and 8, then add 96 to -7.

Hope this helped a bit? :)

4 0
3 years ago
Which graph represents the function R(x) = 821+ +4?
Paha777 [63]

Answer:

answer is i

Step-by-step explanation:

6 0
3 years ago
SOMEONE PLZ HELP ME!!!!!<br> Explain how you find a unit rate when given a rate.
vaieri [72.5K]

Answer:

when giving a rate you need to get a unit rate which is a fraction with a 1 on top then you would divide by the same number as the number if you have 6/6 then divide by 6

Step-by-step explanation:

3 0
3 years ago
A scientist claims that 4% of viruses are airborne. If the scientist is accurate, what is the probability that the proportion of
oksano4ka [1.4K]

Answer:

The probability that the proportion of airborne viruses in a sample of 662 viruses would be greater than 6%=0.00427

Step-by-step explanation:

We are given that

\mu_{\hat{p}}=p=4%=0.04

n=662

We have to find the probability that the proportion of airborne viruses in a sample of 662 viruses would be greater than 6%.

q=1-p=1-0.04=0.96

\sigma_{\hat{p}}=\sqrt{p(1-p)/n}

\sigma_{\hat{p}}=\sqrt{\frac{0.04(1-0.04)}{662}}

\sigma_{\hat{p}}=0.0076

Now,

P(\hat{p}>0.06)=1-P(\hat{p}

=1-P(\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}

=1-P(Z

=1-0.99573

P(\hat{p}>0.06)=0.00427

Hence, the probability that the proportion of airborne viruses in a sample of 662 viruses would be greater than 6%=0.00427

7 0
3 years ago
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