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3 years ago

(2^2xy^3)^5/(x^6y)^6(3yx^2)^10

Mathematics

1 answer:

emmasim [6.3K]3 years ago

8 0

Answer:

do you need to simplify or solve for x?

Step-by-step explanation:

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During 7 1/2 months of hibernation, a black bear experienced a weight loss of 64.4 pounds.

Yakvenalex [24]

Answer:

8.6

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

The median of a set of consecutive odd integers is 138. If the greatest integer in the set is 145, what is the least integer in

garri49 [273]

Answer:

131

Step-by-step explanation:

Given that:

Median of a consecutive odd integers = 138

Greatest integer In dataset = 145

Least integer in dataset =

Since the integers are consecutive ;

Number of integers to greatest integer value : Greatest integer - median value = 145 - 138 = 7

Hence, 7 integers before the median value should give us the least integer value ;

138 - 7 = 131

CHECK:

131,132,133,134,135,136,137,138,139,140,141,142,143,144,145

Median = (n+1)/2 th term

Median = (15+1)/2 = 8th term = 138

Highest = 145

Least = 131

7 0

3 years ago

Is this the right answer??

blsea [12.9K]

Answer:

$135.01

Step-by-step explanation:

A mark up of 100% means that the price was doubled.

58.7*2=117.4

To add sales tax to a price (p) you need to do p+(sales tax * p)

117.4+.15(117.4)

135.01

8 0

3 years ago

Find the nth term of -5,-2,3,10,19

mixer [17]

Answer:

PLace them one by one from big number to small and the middle number is the answer.

Step-by-step explanation:

7 0

3 years ago

In a random sample of 80 teenagers, the average number of texts handled in one day is 50. The 96% confidence interval for the me

Naily [24]

Answer:

Standard deviation of the sample = 17.421

Step-by-step explanation:

We are given that in a random sample of 80 teenagers, the average number of texts handled in one day is 50.

Also, the 96% confidence interval for the mean number of texts handled by teens daily is given as 46 to 54.

So, sample mean, $xbar$ = 50 and Sample size, n = 80

Let sample standard deviation be s.

96% confidence interval for the mean number of texts, $\mu$ is given by ;

96% confidence interval for $\mu$ = $xbar \pm 2.0537*\frac{s}{\sqrt{n} }$

[46 , 54] = $50 \pm 2.0537*\frac{s}{\sqrt{80} }$

Since lower bound of confidence interval = 46

So, 46 = $50 - 2.0537*\frac{s}{\sqrt{80} }$

50 - 46 = $2.0537*\frac{s}{\sqrt{80} }$

s = $\frac{4*\sqrt{80} }{2.0537}$ = 17.421

Therefore, standard deviation of the sample is 17.421 .

3 0

4 years ago