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3 years ago

Which point is a solution to the inequality shown in this graph

Mathematics

2 answers:

astra-53 [7]3 years ago

8 0

Answer:

D: (1,5) Apex

Step-by-step explanation:

Stels [109]3 years ago

8 0

Answer:

Option D. (1, 5)

Step-by-step explanation:

An inequality has been graphed in the given question.

The shaded part upwards the dotted line shows that the solution of the inequality lies in this region.

Now we will consider those points of the options as a solution which lie in the shaded region.

Option A.(3, -1)

Since line is in dotted form therefore points lying on the line are not the solution.

Option B. (5, -5)

Since this point lie below the shaded region so it's not the solution.

Option C. (-3, -3)

This point is again on the dotted line so it's not the solution.

Option D. (1, 5)

This is the correct option because this point lies in the shaded region above the dotted line.

You might be interested in

Area of circle with diameter 14cm is

Alisiya [41]

Answer:

A≈153.94cm²

Step-by-step explanation:

A=πr2

d=2r

Solving forA

A=1

4πd2=1

4·π·142≈153.93804cm²

4 0

3 years ago

Sacramento County high school seniors have an average SAT score of 1,020. From a random sample of 144 Sacramento High School stu

kari74 [83]

Answer:

We need to conduct a hypothesis in order to check if the high school students are representative of the overall population (or if the true mean is 1020 or not), the system of hypothesis would be:

Null hypothesis: $\mu = 1020$

Alternative hypothesis: $\mu \neq 1020$

Step-by-step explanation:

Previous concepts and data given

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".

$\bar X=1100$ represent the sample mean

$s=144$ represent the sample standard deviation

n=144 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the high school students are representative of the overall population (or if the true mean is 1020 or not), the system of hypothesis would be:

Null hypothesis: $\mu = 1020$

Alternative hypothesis: $\mu \neq 1020$

If we analyze the size for the sample is > 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{1100-1020}{\frac{144}{\sqrt{144}}}=6.67$

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=144-1 =143$

Then since is a two sided test the p value would be:

$p_v =2*P(t_{144}>6.67)=2.59x10^{-10}$

Conclusion

If we compare the p value and a significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the population mean is significant different from 1020 at 5% of significance.

6 0

3 years ago

In a survey of 100 students of Surkhet, 65 like to play football, 55 like

Katyanochek1 [597]

Answer:

15 students do not like either sport

Step-by-step explanation:

See the attached image.

Step 1: put 35 in the "like to play both" (intersection) area.

Step 2: subtract those 35 from the 65 football players. 65 - 35 = 30 players like football but not cricket. Put 35 in the football but not cricket area.

Step 3: subtract 35 from the 55 cricket players. 55 - 35 = 20 players like cricket but not football. Put 20 in the cricket but not football area.

Step 4: Add the 30 + 35 + 20 = 85 together. So 85 players are accounted for as players of one or more sports.

Step 5: 100 - 85 = 15 must be the number of students who did not like either sport. Put 15 in the area outside both circles.