PLEASE HELP ME!?!?!

Norbert’s Nursery is having a sale. Flats of flowers are priced as marked, including tax. Jake buys 2 flats of petunias at $5.25

UkoKoshka [18]

Answer:

I am a student. Jake should receive 8.15 dollars in change. Please pick me brainliest! I am new to this website and I wanna help.

Step-by-step explanation:

4 0

3 years ago

Chucky grabbed 11 1111 items in the grocery store that each had a different price and had a mean cost of about $ 4.44 $4.44dolla

maria [59]

Question:

Chucky grabbed 11 items in the grocery store that each had a different price and had a mean cost of about $4.44. On his way to the register, he gave in to an impulse to add a 12th item: an entire wheel of cheese that cost $39.99.

How will adding the wheel of cheese affect the mean and median?

Answer:

There will be a big difference in the mean when the new item is added

Step-by-step explanation:

Given

$Items = 11$

$Mean = \$4.44$

Before we solve further, we need to first calculate the total amount of the 11 items.

$Mean = \frac{Total}{Items}$

Make Total the Subject of formula

$Total = Mean * Items$

$Total = \$4.44 * 11$

$Total = \$48.84$

When the 12th item of $39.99 is added, the new mean becomes.

$New\ Mean = \frac{Total + \$39.99}{11 + 1}$

$New\ Mean = \frac{\$48.84 + \$39.99}{11 + 1}$

$New\ Mean = \frac{\$88.83}{12}$

$New\ Mean = \$7.4025$

By comparing this the old mean, we can see a huge increment between $4.44 and $7.4025

This means that there will be a big difference in the mean when the new item is added.

For the Median:

The old mean shows that the prices of the 11 items is within a small range from $4.44.

So, when the new item is added the median will only change a little bit.

In other words, the median value will only change a little bit.

7 0

3 years ago

Read 2 more answers

A factory makes boxes with length to width ratios of 6 inches to 1inch what should the width of a box with a length of 30 inches

vlada-n [284]

The width of the new enlarged box should be 5 inches. 6/1=30/× Then cross multiply so... 6 time x=6x and 30 times 1=30.(30=6x) Finally, 30 divided by 6 equals 5.

3 0

4 years ago

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The fracture strength of tempered glass averages 14 (measured in thousands ofpounds per square inch) and has standard deviation

jarptica [38.1K]

Answer:

a) 0.62% probability that theaverage fracture strength of 100 randomly selected pieces of this glass exceeds 14.5

b) The interval that includes, with probability 0.95, the average fracture strength of 100 randomly selected pieces of this glass is between 13.61 and 14.39 thousands of pounds per square inch

Step-by-step explanation:

To solve question a, we need to understand the normal probability distribution and the central limit theorem.

To solve question b, we find a normal confidence interval.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

a. What is the probability that theaverage fracture strength of 100 randomly selected pieces of this glass exceeds 14.5?

We have that $\mu = 14, \sigma = 2, n = 100, s = \frac{2}{\sqrt{100}} = 0.2$

This probability is 1 subtracted by the pvalue of Z when X = 14.5. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{14.5 - 14}{0.2}$

$Z = 2.5$

$Z = 2.5$ has a pvalue of 0.9938

1 - 0.9938 = 0.0062

0.62% probability that theaverage fracture strength of 100 randomly selected pieces of this glass exceeds 14.5

b. Find an interval that includes, with probability 0.95, the average fracture strength of 100 randomly selected pieces of this glass.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{2}{\sqrt{100}} = 0.39$

The lower end of the interval is the sample mean subtracted by M. So it is 14 - 0.39 = 13.61

The upper end of the interval is the sample mean added to M. So it is 14 + 0.39 = 14.39

The interval that includes, with probability 0.95, the average fracture strength of 100 randomly selected pieces of this glass is between 13.61 and 14.39 thousands of pounds per square inch

4 0

3 years ago

What is the x-coordinate of the point that divides the

fredd [130]

Answer:

-2

Step-by-step explanation:

In the picture attached, the problem is shown.

The formula you have to use is:

x = m/(m+n)(x2 - x1) + x1

where x1 is the x-coordinate of the origin of the line segment (point J here), x2 is the x-coordinate of the end of the line segment (point K here), and m and n are from the ratio m:n.

In this case:

x1 = -6

x2 = 8

ratio 2:5 means m = 2 and n = 5.

Replacing them into the equation, we get:

x = 2/(2+5)(8 - (-6)) + -6

x = -2

8 0

3 years ago

PLEASE HELP ME!?!?!​

PLEASE HELP ME!?!?!