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2 years ago

I got the answer D is this correct? will vote brainliest

Mathematics

1 answer:

Marina86 [1]2 years ago

4 0

Answer:

yes its a correct answer hope it will help u.......

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The product of a number and four, increased by 5.

german

I’m pretty sure the answer is 4x+5

3 0

1 year ago

Read 2 more answers

Your history teachers is going to give you a 92 point assessment. There are 2 point questions and 3 point questions. If there is

Temka [501]

Answer:

Step-by-step explanation:

x = 2 point questions, y = 3 point questions

x + y = 40

2x + 3y = 92

Solving this we get x = 28, y = 12 so the answer is 28.

4 0

3 years ago

Read 2 more answers

If 13 = 4.5 -5k what is the value of k

USPshnik [31]

Answer:

K = negative 1.7

Step-by-step explanation:

4.5- 13 = -8.5

-8.5/5 = -1.7

4.5 - 5(-1.7) = 13

Hope this helped

7 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

Suppose you have a cache of radium, which has a half-life of approximately 1590 years. how long would you have to wait for 1/5 o

lubasha [3.4K]

<span>You would have to wait 318 years for 1/5 (20%) of the Radium to disappear. 1590 is the point where half of the Radium would decay, so if you take 20% of that, it would equate out to 318 years. Radium has a long half life!</span>

4 0

3 years ago