Find the value of the question

Can someone help me with this plz I am so confused??

ruslelena [56]

F= -1 2/3 or 5/3

In order to find the answer all you have to do is multiply both sides by -4/3.

-3/4f × -4/3 = f

-4/3 × 5/4 = -20/12

-20/12 simplified is -5/3 or -1 2/3

6 0

2 years ago

Read 2 more answers

The area of the base of the regular quadrilateral pyramid is 36 cm2 and the area of a lateral face is 48 cm2. Find: Surface area

Thepotemich [5.8K]

Surface Area (SA) = Base Area (Ab) + Lateral Area (AL)

Base Area = 36cm^2
Lateral Face Area = 48cm^2

*If this is a quadrilateral pyramid, then it has 4 lateral faces, so just multiply the "one" lateral face by 4.

Lateral Area (AL) = 4 x Lateral Face Area
-> AL = 4 x 48 = 192 cm^2

So:

SA = Ab + AL
-> SA = 36 + 192
-> Surface Area = 228 cm^2

3 0

2 years ago

Which statement is most likely to be true for this distribution

Nady [450]

Answer:

???idkv

Step-by-step explanation:

5 0

2 years ago

A marketing manager samples 150 people and finds that 87 of them have made a purchase on the internet within the past month. a.

expeople1 [14]

Answer:

a) $\hat p = \frac{87}{150}= 0.58$

b) We assume for this case a confidence level of 95%

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.58(1-0.58)}{(\frac{0.03}{1.96})^2}=1039.793$

And rounded up we have that n=1040

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".

The population proportion have the following distribution

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

Solution to the problem

Part a

For this case the estimated proportion of people who have made a purchase on the internet within the past month is given by:

$\hat p = \frac{87}{150}= 0.58$

Part b

We assume for this case a confidence level of 95%

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 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by: