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zalisa [80]
3 years ago
11

Answer the following question

Mathematics
1 answer:
marissa [1.9K]3 years ago
4 0
<h3>Answers: Choice A and Choice B</h3>

Explanation:

The 4 in the denominator of the original fraction is the same as 4^1

Then use the rule that (a^b)/(a^c) = a^(b-c). That means (4^32)/(4^1) = 4^(32-1) = 4^31 pointing us to choice B as one of the answers

Then notice how (8^31)/(2^31) = (8/2)^31 = 4^31, meaning choice A is the other answer. The rule I used is (a/b)^c = (a^c)/(b^c).

Choice C is not equivalent to the original expression because 4^4*4^8 = 4^(4+8) = 4^12, which is significantly smaller than 4^31.

Choice D can be ruled out as well because 4^33 is larger than 4^31.

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I think it’s A. If not, then B.
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Ten years ago the father was six times as old as his daughter. After ten years, he will be twice as old as his father . Determin
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Let

10years ago

  • Daughter's age=x
  • Father's age=6x

10years after

  • Daughter's age=x+20
  • Father's age=6x+20

ATQ

\\ \tt\hookrightarrow 6x+20=2(x+20)

\\ \tt\hookrightarrow 6x+20=2x+40

\\ \tt\hookrightarrow 4x=20

\\ \tt\hookrightarrow x=5

  • Father's present age=6x+10=6(5)+10=30+10=40yrs
  • Daughter's present age=x+10=5+10=15yrs
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3 years ago
Read 2 more answers
PLEASE HELP!!! <br>What is the distance, in units, between P=(−6,8) and Q=(9,8)?
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Answer:

15

Step-by-step explanation:

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3 years ago
For his phone service, David pays a monthly fee of $23, and he pays an additional $0.07 per minute of use. The least he has been
Sergeu [11.5K]

Answer:

1019/minutes

Step-by-step explanation:

$94.33 - $23 = $71.33

$71.33 ÷ 0.07 = 1019/mins

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4 0
3 years ago
A certain type of automobile battery is known to last an average of 1110 days with a standard deviation of 50 days. If 100 of th
GalinKa [24]

Answer:

a. P(1104<X[bar]<1110) =  0.3849

b. P(X[bar]>1116) = 0.1151

c. P(X[bar]<940) = 0

Step-by-step explanation:

Hello!

Your study variable is X: "length of life of a battery" (days)

These batteries have a known lifetime average μ: 1110 days and a standard deviation δ:50 days.

To calculate the probabilities I'll apply the Central Limit Theorem, and since the sample is big enough (n>30), approximate the distribution of the sample mean to normal X[bar]≈N(μ;δ²/n) and use the standardization to the Z-distribution to calculate each probability.

a. The average is between 1104 and 1110

P(1104<X[bar]<1110) = P(X[bar]<1110) - P(X[bar]<1104)

= P(Z<(1110-1110)/(50/√100) - P(Z<(1104-1110)/(50/√100))

=P(Z<0) - P(Z<-1.2) = 0.5 - 0.1151 = 0.3849

b. The average is greater than 1116

P(X[bar]>1116) = 1 - P(X[bar]≤1116) = 1 - P(Z≤(1116-1110)/(50/√100))

=1 - P(Z≤1.2) = 1 - 0.8849 = 0.1151

Since the Z table accumulates from left, this means, the probability values it gives are P(Z), to calculate the values ​​above a certain number, you have to subtract from 1 (the highest possible cumulative probability value) all the cumulative probability that is below that number.

c. The average is less than 940.

P(X[bar]<940) = P(Z<(940-1110)/(50/√100)) = P(Z<-34) = 0

I hope you have a SUPER day!

5 0
3 years ago
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