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3 years ago

Identify the pair of exponents that would make this statement FALSE

Mathematics

1 answer:

Lubov Fominskaja [6]3 years ago

8 0

Answer:

the right one is b

Step-by-step explanation:

because if you add all of the other oned it would be 24

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Please answer this question ‍♀️

krek1111 [17]

Answer:

sum is 1,2,4,8,16,32,64,128,256,512

5 0

2 years ago

The number of "destination weddings" has skyrocketed in recent years. For example, many couples are opting to have their wedding

melamori03 [73]

Answer:

We conclude that the mean wedding cost is less than $30,000 as advertised.

Step-by-step explanation:

We are given the following data set:(in thousands)

29100, 28500, 28800, 29400, 29800, 29800, 30100, 30600

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{236100}{8} = 29512.5$

Sum of squares of differences = 3408750

$S.D = \sqrt{\frac{3408750}{7}} = 697.82$

Population mean, μ = $30,000

Sample mean, $\bar{x}$ = $29512.5

Sample size, n = 8

Alpha, α = 0.05

Sample standard deviation, s = $ 697.82

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 30000\text{ dollars}\\H_A: \mu < 30000\text{ dollars}$ We use one-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{29512.5 - 30000}{\frac{697.82}{\sqrt{8}} } = -1.975$

Now,

$t_{critical} \text{ at 0.05 level of significance, 7 degree of freedom } = -1.894$

Since,

$t_{stat} < t_{critical}$

We fail to accept the null hypothesis and reject it.

We conclude that the mean wedding cost is less than $30,000 as advertised.

8 0

3 years ago

The answer to these please!

igomit [66]

$\sqrt[4]{(4a^{2})^{6}} = \sqrt[4]{4096a^{12}} = 8a^{3}$

$\sqrt[7]{(b^{\frac{5}{3}})^{8}} = \sqrt[7]{b^\frac{40}{3}} = b^{\frac{40}{21}} = \sqrt[21]{b^{40}} = \sqrt[21]{b^{21} * b^{19}} = \sqrt[21]{b^{21}} \sqrt[21]{b^{19}} = b\sqrt[21]{b^{19}}$

$\sqrt[3]{\frac{c^{9}}{c^{3}}} = \sqrt[3]{c^{6}} = c^{2}$

$\sqrt[3]{27d^{6} * 8d^{-4}} = \sqrt[3]{27d^{6} * \frac{8}{d^{4}}} = \sqrt[3]{216d^{2}} = 6\sqrt[3]{d^{2}} = 6d^{\frac{2}{3}}$

4 0

3 years ago

Read 2 more answers

Find the value of a - b when a = 3 and b = -2.

lubasha [3.4K]

Answer:

Step-by-step explanation:

a = 3; b = -2

a - b = 3 - (-2) = 3 + 2 = 5

3 0

2 years ago

Jon’s weight loss for each week of the month is 4 lbs., 3.5 lbs. and 2 lbs. He gained 4.5 lbs. the last week. If Jon originally

nekit [7.7K]

Answer: 148 lbs

first you do 4+3.5+2=9.5 then you do 9.5-4.5=5 so then you do 153-5 to get your answer 148 lbs

i hope this helps :)

4 0

3 years ago