All categories

3 years ago

Estimate the sum to the nearest hundreds 166+427

Mathematics

2 answers:

quester [9]3 years ago

5 0

The answer to your question is 600

mr_godi [17]3 years ago

4 0

166 estimated to the nearest hundreds would be 200 (6 is greater than 5, and so you round up. & since you are rounding to the nearest hundred, you have to look at the digit in the tens place)

427 estimated to the nearest hundreds would be 400

400 + 200 = 600

your estimated sum is 600

hope this helps

You might be interested in

100 POINTS!!! VERY URGENT

Marta_Voda [28]

Answer:

D) (f o g)(x) = 10x² - 60x + 93

Step-by-step explanation:

f(x) = 10x² + 3

g(x) = x - 3

⇒ (f o g)(x)

⇒ f(x - 3)

⇒ f(x - 3) = 10(x - 3)² + 3

⇒ f(x - 3) = 10(x² - 6x + 9) + 3

⇒ f(x - 3) = 10x² - 60x + 90 + 3

⇒ (f o g)(x) = 10x² - 60x + 93

6 0

2 years ago

Read 2 more answers

The owner of a football team claims that the average attendance at games is over 523, and he is therefore justified in moving th

USPshnik [31]

Answer:

C) There is not sufficient evidence to support the claim that the mean attendance is greater than 523.

Step-by-step explanation:

Let μ be the the average attendance at games of the football team

The claim: the average attendance at games is over 523

Null and alternative hypotheses are:

$H_{0}$ : μ=523

$H_{a}$ : μ>523

The conclusion is failure to reject the null hypothesis.

This means that test statistic is lower than critical value. Therefore it is not significant, there is no significant evidence to accept the alternative hypothesis.

That is no significant evidence that the average attendance at games of the football team is greater than 523.

7 0

3 years ago

Donatello Co. has identified an activity cost pool to which it has allocated estimated overhead of $9,600,000. It has determined

ahrayia [7]

Answer:

option (a) Widgets $2,400,000, Gadgets $1,800,000, Targets $5,400,000

Step-by-step explanation:

Data provided in the question:

Allocated estimated overhead = $9,600,000

Expected use of cost drivers for the activity = 800,000 inspections.

Inspections required by Widgets = 200,000

Inspections required by Gadgets = 150,000

Inspections required by Targets = 450,000

Therefore,

Total overhead per activity = $\frac{\textup{estimated overhead}}{\textup{Total activity}}$

= $\frac{\textup{9,600,000}}{\textup{800,000 }}$

= $12 per activity

Thus,

Overhead allocated to Widgets = 200,000 × 12 = $2,400,000

Overhead allocated to Gadgets = 150,000 × 12 = $1,800,000

Overhead allocated to Target = 450,000 × 12 = $5,400,000

Hence, The correct answer is option (a)

8 0

2 years ago

In a game, there are 12 identical balls of which seven are red and five are green.

Alex

Answer:

its number 2 and if its a mutable answers writ 3 also

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