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3 years ago

If x3 = -27, then what is x?

Mathematics

1 answer:

inn [45]3 years ago

5 0

Answer:

x = -3

Step-by-step explanation:

x^3 = -27

Take the cube root of each side

(x^3) ^1/3 = -27 ^1/3

x = -3

We know that -3*-3*-3 = -27 so this checks

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A 276-inch board is cut into two pieces. One piece is five times the length of the other. Find the lengths of the two pieces

timurjin [86]

The smallest is 1/(5+1) = 1/6 of the total length, so is (276 in)/6 = 46 in.
The largest is 5*46 in = 230 in.

The two pieces are 46 in and 230 in.

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3 years ago

Choose the estimate for 58÷589

Elina [12.6K]

10 is the estimate because you have 58, and you put it in 589 10 times.

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3 years ago

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A company made a storage container with sheet steel walls. The container was in the shape of a rectangular prism, as shown below

Alja [10]

Answer:

The sheet steel costed $22.41 per square meter

Step-by-step explanation:

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3 years ago

Name the property used in: (5 + 2) + n = 7 + n

rjkz [21]

Additive indentity
$(5 + 2) + n$
n = 1
a variable is always a number itself remember 7+n
5+2 = 7

5 0

4 years ago

Suppose that the mean water hardness of lakes in Kansas is 425 mg/L and these values tend to follow a normal distribution. A lim

tino4ka555 [31]

Answer:

The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

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.

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

$Mean =\displaystyle\frac{10959}{25} =438.36$

Sum of squares of differences = 175413.76

$S.D = \sqrt{\dfrac{175413.76}{24}} = 85.49$

Population mean, μ = 425 mg/L

Sample mean, $\bar{x}$ = 438.36

Sample size, n = 25

Alpha, α = 0.05

Sample standard deviation, s = 85.49

First, we design the null and the alternate hypothesis

$H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}$

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

$t_{stat} = \displaystyle\frac{438.36 - 425}{\frac{85.49}{\sqrt{25}} } = 0.7813$

Now, $t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = -1.7108$

Since,

The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

6 0

3 years ago