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3 years ago

9,000 + 40 + 0.1 + 0.008 I need to put it in standard form. Please help!

Mathematics

1 answer:

Andre45 [30]3 years ago

8 0

Answer:

-12

Step-by-step explanation:

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An enclosed gas exerts a pressure P on the walls of a container. This pressure is directly proportional to the temperature T of

tatyana61 [14]

Don't know if i'm right but I got 1/60

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3 years ago

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If 5 students had to share 8 brownies how much would each student get

muminat

Each student would get 1.6 brownies

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3 years ago

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14. Use the given degree of confidence and Sample data to construct a confidence interval for the population mean. Assume that t

nirvana33 [79]

Step-by-step explanation:

Confidence interval = mean ± margin of error

CI = μ ± ME

The mean is μ = 8.7.

Margin of error = critical value × standard error

ME = CV × SE

At 95% confidence and 9 degrees of freedom, CV = 2.262.

SE = s / √n

SE = 3.3 / √10

SE = 1.04

The margin of error is:

ME = 2.262 × 1.04

ME = 2.36

CI = 8.7 ± 2.36

CI = (6.34, 11.06)

7 0

3 years ago

I only need questions 2 and 3 it’s urgent could someone please help?

marta [7]

Answer:

a = x-intercept(s): ( 3 , 0 )

y-intercept(s):

( 0 , 6 )

B = x-intercept(s):

( 10 , 0 )

y-intercept(s):

( 0 , 4 )

Step-by-step explanation:

3 0

2 years ago

Use the null hypothesis H0 : μ = 98.6, alternative hypothesis Ha: μ < 98.6, and level of significance α = 0.05. 98 99.6 97.8

andrezito [222]

Answer:

The t-score is -1.8432

Step-by-step explanation:

We are given the following in the question:

98, 99.6, 97.8, 97.6, 98.7, 98.4, 98.9, 97.1, 99.2, 97.4, 99.1, 96.9, 98.8, 99.9, 96.8, 97, 98.7, 97.6, 98.7, 98.2

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.

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

$Mean =\displaystyle\frac{1964.4}{20} = 98.22$

Sum of squares of differences = 16.152

$s = \sqrt{\dfrac{16.152}{49}} = 0.922$

Population mean, μ = 98.6

Sample mean, $\bar{x}$ = 98.22

Sample size, n = 20

Sample standard deviation, s = 0.922

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 98.6\\H_A: \mu < 98.6$

Formula:

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

Putting all the values, we have

$t_{stat} = \displaystyle\frac{98.22 - 98.6}{\frac{0.922}{\sqrt{20}} } = -1.8432$

The t-score is -1.8432

7 0

3 years ago