-6x- 17/4 is less than or equal to 79/4<br><br> please please please solve and show work

Please i really need help

Airida [17]

Answer:

-6

Step-by-step explanation:

it's asking when x is equal to -5 what is y equal to?

there's only one point when x=-5 and it's -6

so the point is (-5,-6) and y is equal to -6

hope this helps and makes sense <3

do well babe

6 0

3 years ago

In a survey of 2,000 people who owned a certain type of car, 1,100 said they would buy that type of car again. What percent of t

Mrac [35]

Answer:
The percentage of people surveyed that were satisfied with the car is 55%.
Step-by-step explanation:
The statement indicates that in a survey 1,100 people of 2,000 surveyed said that they would buy the car they have again and from this, you can inferred that they are satisfied with the car. So, to calculate the percentage of people surveyed that were satisfied with the car you have to divide 1,100 by 2,000 that is the total number of people surveyed and then, multiply the result by 100:
1,100/2,000=0.55*100= 55%

Read more on Brainly.com - brainly.com/question/5991333#readmore

7 0

3 years ago

What is the difference of

inysia [295]

Answer:65,

Step-by-step explanation:

8 0

3 years ago

Can someone answer this question please answer it correctly if it’s corect I will mark you brainliest

Sunny_sXe [5.5K]

Answer:

58

Step-by-step explanation:

upper quartile is the median of the upper half of a data set

The median is the middle number which is 37

There are 5 numbers left in the upper half

The middle number or median of the upper half is 58

6 0

3 years ago

Read 2 more answers

23. A certain type of gasoline is supposed to have a mean octane rating of at least 90. Suppose measurements are taken on of 5 r

Nostrana [21]

Answer:

a) $p_v =P(Z$

b) Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

c) $P(\bar X >90)=1-P(Z$

$1-P(Z$

d) For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

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".

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

$\bar X=88.96$ represent the sample mean

$s=1.011$ represent the sample standard deviation

n=5 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 mean is less than 90, the system of hypothesis would be:

Null hypothesis: $\mu \geq 90$

Alternative hypothesis: $\mu < 90$

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:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-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:

$z=\frac{88.96-90}{\frac{0.8}{\sqrt{5}}}=-2.907$

Part a

P-value

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

$df=n-1=5-1 = 4$

Then since is a left tailed test the p value would be:

$p_v =P(Z$

Part b

Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

Part c

We want this probability:

$P(\bar X$

The best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

If we apply this formula to our probability we got this:

$P(\bar X >90)=1-P(Z$

$1-P(Z$

Part d

For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

7 0

3 years ago

-6x- 17/4 is less than or equal to 79/4 please please please solve and show work