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2 years ago

Can some one please explain how to do this? 1.253 x 7.4 =

Mathematics

1 answer:

drek231 [11]2 years ago

3 0

Answer:

9.2722

Step-by-step explanation:

7.400

*1.252

_____

9.2722

Its simple, plz rate 5 str, thx, and brailiest

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Please identify the common angles and sides or if there are non. Explain all work please!

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Answers:

Common angles: ΔBGC and ΔGDC

Common sides: ΔGE and ΔAG

A common angle is a triangle that exists in two or both of the triangles.

Here, the common angles should be ΔBGC and ΔGDC.

These two angles are triangles, and they consist in both triangles.

A common side is when two angles have one vertex in the same area.

Here, we see that the common sides are ΔGE, because they intersect.

Another common side we see here is ΔAG, because they also intersect.

Hope this answer helps you!

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Answer:

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Step-by-step explanation:

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2 years ago

Of 575 broiler chickens purchased from various kinds of food stores in different regions of a country and tested for types of ba

chubhunter [2.5K]

Answer:

a) $0.68 - 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.630$

$0.68 + 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.730$

And the 99% confidence interval would be given (0.630;0.730).

b) We are 99% confident that the true proportion of infected chickens are between 0.63 (63%) and 0.73 (73%)

c) For this case the answer would be No. Since all the required assumptions in order to construct the confidence interval are satisfied, so then the results can be assumed for all the population

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".

$p$ represent the real population proportion of interest

$\hat p =0.68$ represent the estimated proportion

n=575 is the sample size required (variable of interest)

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

The population proportion have the following distribution

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

Part a

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$0.68 - 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.630$

$0.68 + 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.730$

And the 99% confidence interval would be given (0.630;0.730).

Part b

We are 99% confident that the true proportion of infected chickens are between 0.63 (63%) and 0.73 (73%)

Part c

For this case the answer would be No. Since all the required assumptions in order to construct the confidence interval are satisfied, so then the results can be assumed for all the population

4 0

3 years ago