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3 years ago

Help please will give brainllst to the right answer

Mathematics

2 answers:

svet-max [94.6K]3 years ago

7 0

Answer :

Your answer will be 7

stich3 [128]3 years ago

3 0

Answer:

Step-by-step explanation:

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Which lists all the integer solutions of the inequality |x| < 3?

Alexus [3.1K]

Simple the answer is c 100% positive

5 0

3 years ago

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Perimeter of rectangle floor is 80 feet. Find the dimention of the floor if the length is four times the width

erik [133]

Answer:

width = 8 ft & length = 32 ft.

Step-by-step explanation:

let the width be x and length be 4x.

we know, Perimeter = 2(l+b)

now,

2(4x + x) = 80

10x = 80

x = 80/10 = 8

x = 8

therefore,

width = x = 8 ft

length = 4x = 4 × 8 = 32 ft.

hope this helps you !

6 0

2 years ago

A population of insects grows exponentially, as shown in the table. Suppose the increase in population continues at the same rat

Ivanshal [37]

We are told that a population of insects grows exponentially and we are given a table of data about insect population growth. We are asked to find population of insects at the end of week 11.

The initial insect population is 20 and at the end of 1st week population increases to 30.

Let us find growth percentage of insect population,

$\text{Growth percentage}=\frac{\text{Difference}}{Actual} \cdot100$

$\text{Growth percentage}=\frac{\text{30-20}}{20} \cdot100$

$\text{Growth percentage}=\frac{\text{10}}{20} \cdot100$

$\text{Growth percentage}=0.5 \cdot100=50$

We can see that insect population is growing at rate of 50% per week.

Now let us write an exponential function for our population.

$P(w)=20(1+0.50)^{w}$ , where P(w) represents population at the end of w weeks.

Let us substitute w=11 in our function to find insect population at the end of 11 weeks.

$P(11)=20(1+0.50)^{11}$

$P(11)=20(1.50)^{11}$

$P(11)=20\cdot 86.49755859375$

$P(11)=1729.951171875\approx 1730$

Therefore, population of insects at the end of 11th week will be 1730.

7 0

3 years ago

Find f(-3) if f(x) = x^2.

kogti [31]

Hopefully that’s help u :)

3 0

2 years ago

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Researchers have studied the role that the age or workers has in determining the hours per month spent on personal tasks. A samp

ehidna [41]

Answer:

$4.31-1.96\frac{0.82}{\sqrt{677}}=4.248$

$4.31+1.96\frac{0.82}{\sqrt{677}}=4.372$

So on this case the 95% confidence interval would be given by (4.248;4.372)

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".

$\bar X= 4.31$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=0.82 represent the sample standard deviation

n=677 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=677-1=676$

With this sample size is important to remark that we can use the normal distribution as an approximation since the sample size is large enough.

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,676)".And we see that $t_{\alpha/2}=1.96$

Now we have everything in order to replace into formula (1):

$4.31-1.96\frac{0.82}{\sqrt{677}}=4.248$

$4.31+1.96\frac{0.82}{\sqrt{677}}=4.372$

So on this case the 95% confidence interval would be given by (4.248;4.372)

8 0

3 years ago