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artcher [175]
4 years ago
6

What is 1,000,000,000 divided by 1,000,000,000,000,000,000,000 plus 300 400

Mathematics
2 answers:
Nesterboy [21]4 years ago
8 0

Answer:

700

Step-by-step explanation:

iris [78.8K]4 years ago
3 0

Answer:

700

Step-by-step explanation:

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Could anyone help me with this scatter plot question? Thanks :)
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2 + 4x - 4 = 0 which gives x =
ExtremeBDS [4]

Answer:

x =  \frac{1}{2}

Step-by-step explanation:

2 + 4x - 4 = 0 \\ 2 + 4x = 0 + 4 \\ 2 + 4x = 4 \\ 4x = 4 - 2 \\ 4x = 2 \\  \frac{4x}{4}  =  \frac{2}{4}  \\ x =  \frac{1}{2}

7 0
3 years ago
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A family wants to remodel a house and spend at most $135,000 total for the entire project. a)If the builder wants to make a 15%
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3 years ago
What is the volume, in cubic ft, of a rectangular prism with a height of 13ft, a width of 15ft, and a length of 9ft?
Norma-Jean [14]

Answer:

1755 ft³

Step-by-step explanation:

Volume of a rectangular prism = base area x height

Base area = length x width

15 ft x 9 ft = 135 ft²

Volume =  135 ft² x 13 ft = 1755 ft³

5 0
3 years ago
Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below. Find the 93% conf
kobusy [5.1K]

Answer:

(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}

And replacing we got:

(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903

(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097

And the confidence interval for the difference of means would be given by:

0.2903 \leq \mu_1 -\mu_2 \leq 0.5097

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

We have the following data given:

\bar X_1 = 9.2 , \bar X_2 = 8.8, \sigma_1= 0.3, n_1 = 27, \sigma_2 = 0.1, n_2 = 30

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 93% of confidence, our significance level would be given by \alpha=1-0.93=0.07 and \alpha/2 =0.035. And the critical value would be given by:  

z_{\alpha/2}=-1.811, z_{1-\alpha/2}=1.811  

The confidence interval is given by:

(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}

And replacing we got:

(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903

(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097

And the confidence interval for the difference of means would be given by:

0.2903 \leq \mu_1 -\mu_2 \leq 0.5097

4 0
3 years ago
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