All categories

3 years ago

600,000 +80,000+10 to standard form

2 answers:

8 0

To solve this problem to standard form means adding all those numbers together.

600,000+80,000= 680,000
680,000+10= 680,010

680,010 is your answer

11Alexandr11 [23.1K]3 years ago

3 0

68,010 that is the answer when you add of them up

You might be interested in

Well this is the question. The Scoop website charges $29 for a four line classified ad. Each additional line costs $3.45. For an

ohaa [14]

Explanation:

The cost for the ad can be calculated as:

Cost = $29 + 3.45(3) + $20

Cost = $59.35

Because the ad has 7 lines, 4 that are include in the initial

4 0

1 year ago

Find the solution for the following: Three backpackers cooked rice for dinner. The first one gave 400g of rice and the second 20

victus00 [196]

Answer:

First backpacker should receive $4 and the second one $2

Step-by-step explanation:

notice that the total amount of rice for the three is 600 g.

Then, the first one that gave 400 g contributed 400 out of 600, that is 400/600 = 2/3

The second one contributed 200 out of 600, that is 200/600 = 1/3

then the first one should receive 2/3 of the $6 = (2/3) x 6 = $4

and the second one should receive 1/3 of the %6 = (1/3) x 6 = $2

6 0

4 years ago

A study was recently conducted at a major university to estimate the difference in the proportion of business school graduates w

sveta [45]

Answer:

$(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412$

$(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318$

And the 95% confidence interval would be given (-0.1412;-0.0318).

We are confident at 95% that the difference between the two proportions is between $-0.1412 \leq p_A -p_B \leq -0.0318$

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".

$p_A$ represent the real population proportion for business

$\hat p_A =\frac{75}{400}=0.1875$ represent the estimated proportion for Business

$n_A=400$ is the sample size required for Business

$p_B$ represent the real population proportion for non Business

$\hat p_B =\frac{137}{500}=0.274$ represent the estimated proportion for non Business

$n_B=500$ is the sample size required for non Business

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412$

$(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318$

And the 95% confidence interval would be given (-0.1412;-0.0318).

We are confident at 95% that the difference between the two proportions is between $-0.1412 \leq p_A -p_B \leq -0.0318$

7 0

3 years ago

Brums [2.3K]

Step-by-step explanation:

to my guess the rule of a reflection will be about the x-axis since only the values of y have changed

7 0

3 years ago

The standard deviation for actuary salaries has a standard deviation of $36,730 . You collect a simple random sample of n=36 sal

Yanka [14]

Using the z-distribution, it is found that the lower limit of the 95% confidence interval is of $99,002.

<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

In which:

$\overline{x}$ is the sample mean.

z is the critical value.

n is the sample size.

$\sigma$ is the standard deviation for the population.

In this problem, we have a 95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so the critical value is z = 1.96.

The other parameters are given as follows:

$\overline{x} = 111000, \sigma = 36730, n = 36$

Hence, the lower bound of the interval is:

$\overline{x} - z\frac{\sigma}{\sqrt{n}} = 111000 - 1.96\frac{36730}{\sqrt{36}} = 99002$

The lower limit of the 95% confidence interval is of $99,002.

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

8 0

2 years ago