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zalisa [80]
3 years ago
8

6x6x6x6x6x6 15x15x15 Write the product as a power

Mathematics
2 answers:
Alenkinab [10]3 years ago
7 0

Step-by-step explanation:

Add the amount of the same number together. That will be your power:

15 x 15 x 15 = 15³

6 x 6 x 6 x 6 x 6 x 6 = 6^6

~

Assoli18 [71]3 years ago
7 0
6^6 15^3 I need 20 characters so ignore this
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Help ASAP please please 14 and 15
Yuliya22 [10]
I think 14 is B (idrk) and 15 is definitely C
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3 years ago
Question Help Suppose that the lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a
koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean \mu and standard deviation \sigma, the z-score of a value X is given by:

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

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So \mu = 5656, \sigma = 333.3

(a) What proportion of light bulbs will last more than 6262 ​hours?

The pvalue of the z-score of X = 6262 is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.

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

Z = \frac{6262 - 5656}{333.3}

Z = 1.82

Z = 1.81 has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So

P = 100% - 96.562% = 3.438% of the light bulbs will last more than 6262 hours.

​(b) What proportion of light bulbs will last 5252 hours or​ less?

This is the pvalue of the zscore of X = 5252

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

Z = \frac{5252- 5656}{333.3}

Z = -1.21

Z = -1.21 has a pvalue of .1131. This means that 11.31% of the light bulbs will last 5252 hours or less.

(c) What proportion of light bulbs will last between 5858 and 6262 ​hours?

This is the pvalue of the zscore of X = 6262 subtracted by the pvalue of the zscore X = 5858

For X = 6262, we have that Z = 1.81 with a pvalue of .96562.

For X = 5858

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

Z = \frac{5858- 5656}{333.3}

Z = 0.61

Z = 0.61 has a pvalue of .72907.

So, the proportion of light bulbs that will last between 5858 and 6262 hours is

P = .96562 - .72907 = .23655

23.655% of the light bulbs are going to last between 5858 and 6262 hours.

​(d) What is the probability that a randomly selected light bulb lasts less than 4646 ​hours?

This is the pvalue of the zscore of X = 4646

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

Z = \frac{4646- 5656}{333.3}

Z = -3.03

Z = -3.03 has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.

5 0
3 years ago
Set C is the set of two-digit even numbers greater than 34 that are divisible by 5<br><br> C=
amid [387]
In order for it to be divisible by 5 AND be an even number, it must also be divisible by 2. So you are looking for two-digit numbers greater than 34 and divisible by 10:
C = {40, 50, 60, 70, 80, 90}
8 0
3 years ago
Please help find the area (not cm) i’ll give brainliest
Liono4ka [1.6K]

Answer

4.5 centimeters2

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
How do you simplify 1/5x -3 =4
SCORPION-xisa [38]

Answer:

35

Step-by-step explanation:

1/5x - 3 = 4

first move 3 to the other side with addition

1/5x = 7

then you can divide 7 by 1/5

7/1 • 5/1 (when dividing fractions the second one is always the opposite reciprocal)

=35/1 = 35

5 0
3 years ago
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