All categories

2 years ago

Type the correct answer in the box. Use numerals instead of words. -3n^2+147=0

Mathematics

2 answers:

Ber [7]2 years ago

6 0

Answer:

n=7 or n=-7

Step-by-step explanation:

Angelina_Jolie [31]2 years ago

5 0

Answer:

n=7, n=-7

Step-by-step explanation:

You might be interested in

Which statement is true regarding the graphed functions?

andreev551 [17]

There is no graph, so it is impossible to answer this question. I apologise.

7 0

3 years ago

There are three concerts at lakeside one has 953 as attendance and the other 2 have 1100 as the attendance what is the average o

Vilka [71]

You have to add all three concerts 953+1100+1100 =3,163
And the answer you have divided by 3
The average of people at the concerts were 1,054

3 0

3 years ago

−9,000÷3=−3,000 What does -3,000 tell me

soldier1979 [14.2K]

Answer:

Step-by-step explanation:

SEARCH IT UP!!!!!!!

3 0

3 years ago

5. The superintendent of the local school district claims that the children in her district are brighter, on average, than the g

anygoal [31]

Answer:

We conclude that children in district are brighter, on average, than the general population.

Step-by-step explanation:

We are given the following data set:

105, 109, 115, 112, 124, 115, 103, 110, 125, 99

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.

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

$Mean =\displaystyle\frac{1117}{10} = 111.7$

Sum of squares of differences = 642.1

$S.D = \sqrt{\frac{642.1}{49}} = 8.44$

We are given the following in the question:

Population mean, μ = 106

Sample mean, $\bar{x}$ = 111.7

Sample size, n = 10

Alpha, α = 0.05

Sample standard deviation, s = 8.44

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 106\\H_A: \mu > 106$

We use one-tailed(right) t test to perform this hypothesis.

Formula:

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

Putting all the values, we have

$t_{stat} = \displaystyle\frac{111.7 - 106}{\frac{8.44}{\sqrt{10}} } = 2.135$

Now,

$t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.833$

Since,

$t_{stat} > t_{critical}$

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

We conclude that children in district are brighter, on average, than the general population.

4 0

3 years ago

Need some help with this please!!

beks73 [17]

Answer:

B,C and B,A

Step-by-step explanation:

Line the cordients and go up right,Then left to find your answer.

4 0

2 years ago