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3 years ago

70 is 30% of what number?

Mathematics

2 answers:

ExtremeBDS [4]3 years ago

5 0

Answer: 233.33.

Step-by-step explanation: Consider marking this answer as brainliest if it helped you out.

Afina-wow [57]3 years ago

4 0

Answer:

Step-by-step explanation:

233.33

Hope this helps =)

You might be interested in

What is the unit rate of 128 dog food for 16 dogs

Sauron [17]

Answer:

128 ÷ 16 = 8

so 8 dog food for every dog

7 0

3 years ago

find x if the distance between points L and M is 15 and point M is located in the first quadrant. L=(-6,2) M=(x,2)

aivan3 [116]

Answer:

Therefore,

$x = 9$

Step-by-step explanation:

Given:

Let,

point L( x₁ , y₁) ≡ ( -6 , 2)

point M( x₂ , y₂ )≡ (x , 2)

l(AB) = 15 units (distance between points L and M)

To Find:

x = ?

Solution:

Distance formula between Two points is given as

$l(LM)^{2} = (x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$

Substituting the values we get

$15^{2}=(x--6)^{2}+(2-2)^{2}\\\\225=(x+6)^{2}$

Square Rooting we get

$(x+6)=\pm \sqrt{225}=\pm 15\\\\x+6 = 15\ or\ x+6 = -15\\\\x= 9\ or\ x = -21$

As point M is located in the first quadrant

x coordinate and y coordinate are positive

So x = -21 DISCARDED

Hence,

$x = 9$

Therefore,

$x = 9$

3 0

3 years ago

A government report gives a 99% confidence interval for the proportion of welfare recipients who have been receiving welfare ben

juin [17]

Answer:

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

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

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

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".

Solution to the problem

The population proportion have the following distribution

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

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 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.

$z_{\alpha/2}=1.96$

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.