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3 years ago

The lengths, in centimeters, of nine earthworms are shown below. 3, 4, 5, 5, 6, 7, 8, 9, 10 What is the median of the data?

Mathematics

1 answer:

mel-nik [20]3 years ago

6 0

The median of the problem is 6 because it is in the middle of the number set of the earthworm sizes.

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The length of the base of a rectangular prism is given as x + 4, and the width of the base is x +2. The height of the rectangula

lutik1710 [3]

Answer:

V(x) = (x +4)(x +2)(2x +11)

Step-by-step explanation:

For length L = (x+4), the height is ...

2L +3 = 2(x+4) +3

= 2x +8 +3

= 2x +11

The volume is the product of length, width, and height, so is ...

V = (x +4)(x +2)(2x +11)

8 0

3 years ago

90 - 15 x 12 divided 6 Plz help! I’ll give some points!

PilotLPTM [1.2K]

I believe it’s 900. Yeah, I think the answer is 900.

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3 years ago

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Help meh! <br><br><br> Use digits to write the decimal number seventy-six and five hundredths. :))

Kaylis [27]

Answer:

76.05

Step-by-step explanation:

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3 years ago

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Find the inverse of g(x)=1/3x - 7 and then find its domain and range if you can show steps that would be gladly appreciated than

Schach [20]

Answer:

g-1(x) = 3(x + 7).

Domain is All Real values of x.

Range is All Real values of the function.

Step-by-step explanation:

Let y = 1/3 x - 7

1/3 x = y + 7

Multiply both sides by 3:

x = 3(y + 7)

So the inverse of g(x) = g-1(x) = 3(x + 7).

Domain is All Real values of x.

Range is All Real values of the function.

6 0

3 years ago

A random sample of 100 observations from a quantitative population produced a sample mean of 22.8 and a sample standard deviatio

natima [27]

Answer:

We conclude that the population mean is 24.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 24

Sample mean, $\bar{x}$ = 22.8

Sample size, n = 100

Alpha, α = 0.05

Sample standard deviation, s = 8.3

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 24\\H_A: \mu \neq 24$

We use Two-tailed z test to perform this hypothesis.

Formula:

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

Putting all the values, we have

$z_{stat} = \displaystyle\frac{22.8 - 24}{\frac{8.3}{\sqrt{100}} } = -1.44$

We calculate the p-value with the help of standard z table.

P-value = 0.1498

Since the p-value is greater than the significance level, we accept the null hypothesis. The population mean is 24.

Now, $z_{critical} \text{ at 0.05 level of significance } = \pm 1.96$

Since, the z-statistic lies in the acceptance region which is from -1.96 to +1.96, we accept the null hypothesis and conclude that the population mean is 24.

7 0

3 years ago