Please help i have no idea how to do this

There are between 90 and 115 cars in a parking lot. Exactly one-fourth of them are green. Exactly one-eighth of them have a stic

Schach [20]

There are 104 cars in the parking lot.

According to statement there are between 90 and 115 cars on the lot.

So, {X| 90 < x < 115} (This renders an infinite solution set finite)

AND exactly one eight of them have a sticker on the back, so the total number of cars must be evenly divisible by eight.

X ∈ {96, 104, 112,}

AND exactly one fourth of the cars are green, so the number of cars must be evenly divisible by 4. Here all above written numbers are divisible by 4. So, find the mean to calculate the number of cars in the parking lot.

x = (96+104+112)/3

x = 104

There are 104 cars in the parking lot.

Learn more about ELIMINATION METHOD here brainly.com/question/13729904

#SPJ4

5 0

1 year ago

Apparently what I chose was incorrect. Which ones are correct?

ch4aika [34]

Step-by-step explanation:

numbers one two and three are correct

4 0

3 years ago

Helps me pls what is 283x2492÷8(89)=2838(82x) what is x

Andrei [34K]

Answer:

Hey it’s 283

Step-by-step explanation:

3 0

3 years ago

Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below. Find the 93% conf

kobusy [5.1K]

Answer:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}$

And replacing we got:

$(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903$

$(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097$

And the confidence interval for the difference of means would be given by:

$0.2903 \leq \mu_1 -\mu_2 \leq 0.5097$

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".

We have the following data given:

$\bar X_1 = 9.2 , \bar X_2 = 8.8, \sigma_1= 0.3, n_1 = 27, \sigma_2 = 0.1, n_2 = 30$

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 93% of confidence, our significance level would be given by $\alpha=1-0.93=0.07$ and $\alpha/2 =0.035$ . And the critical value would be given by: