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n200080 [17]
3 years ago
8

Which exponential expression is equal to 2−5·28

Mathematics
1 answer:
gizmo_the_mogwai [7]3 years ago
3 0

Answer:

Impossible

Step-by-step explanation:

Solve the answer 2-5*28 = -138

This would have no answer since an exponential expression can't equal an negative unless it's a negative answer, but in this case, there is no answer that would get you -138.

You might be interested in
Miguel tells his teacher1/5 is the same as 20%. Which best justifies Miguel’s answer? a.5 goes into 100 twenty times, so 20% is
svetoff [14.1K]

Answer:

C. 5 goes into 100 twenty times, and 1 times 20 is 20

Step-by-step explanation:

Since, we know that when we multiply both numerator and denominator of a fraction by a same number then we obtain an equivalent fraction,

Here, the given fraction,

\frac{1}{5}

By the above statement,

\frac{1}{5}=\frac{1\times 20}{5\times 20}=\frac{20}{100}

Now,

a\%=\frac{a}{100}

\implies \frac{20}{100}=20\%

Hence,

\frac{1}{5}=20\%

Option C is correct.

3 0
2 years ago
Read 2 more answers
(15 - 6) - 3 = 15 - (6 - 3). is julie's wrote equation sense or nonsense? do you think the associative property works for subtra
never [62]

nonsense because, 6 ≠ 12

Step-by-step explanation:

(15 - 6) - 3 = 15 - (6 - 3)

you gotta to the parentheses first.

9 - 3 = 15 - 3

6 = 12

6 ≠ 12

5 0
3 years ago
What is 1,500mm + 15mm + 5m =
storchak [24]
1,520
add 1,500 plus 15 plus 5
4 0
2 years ago
Suppose the horses in a large stable have a mean weight of 975lbs, and a standard deviation of 52lbs. What is the probability th
Lubov Fominskaja [6]

Answer:

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 975, \sigma = 52, n = 31, s = \frac{52}{\sqrt{31}} = 9.34

What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable?

pvalue of Z when X = 975 + 15 = 990 subtracted by the pvalue of Z when X = 975 - 15 = 960. So

X = 990

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{990 - 975}{9.34}

Z = 1.61

Z = 1.61 has a pvalue of 0.9463

X = 960

Z = \frac{X - \mu}{s}

Z = \frac{960 - 975}{9.34}

Z = -1.61

Z = -1.61 has a pvalue of 0.0537

0.9463 - 0.0537 = 0.8926

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

7 0
3 years ago
Three moons are in the same circular orbit around a planet. The moons are each 115,000 kilometers from the surface of the planet
marshall27 [118]
275,000 km
The moons A, B, C form a right triangle with AC forming a hypotenuse with the diameter of the planet and the distance of both moons from the surface. So add A to the surface, plus the diameter of the planet, plus surface to c.
115000+45000+115000 = 275000
7 0
2 years ago
Read 2 more answers
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