Write 7,447,000 in scientific notation.

Helppppp plsss<br> Ill mark brainliest 20pts

Citrus2011 [14]

A) For both sets A and B, calculating the mean, range, and quartiles are a good way of measuring the center and spread. Using standard deviation may not be the best because we do not know whether the distributions are normal or not.

b) For set A, the lowest value is 63, while the highest is 86. An estimate for the mean, based on the average of these, is 74.5. Most of the 70+ values are below 74.5, so we may guess that the mean will be above the median.
For set B, the lowest is 63, while the maximum is 95. The estimated mean would be 79. But since there are more values on the 80+ and 90+ side, the median is likely to be higher than 79.

c) For set A, the mean is 74.79, while the median is 73, therefore the mean is above the median, and the prediction in part b is correct.

4 0

3 years ago

⚠️HELPP PLEASE⚠️ Find the length of side A

exis [7]

Answer:

pretty sure that's C

Step-by-step explanation:

4 0

3 years ago

2) You can buy 3 apples at the Quick Market for $1.05. You can buy 5 of the

Alexus [3.1K]

Answer:

Quick Market

Step-by-step explanation:

You need to set up the problem in fraction form or 3/1.05 and 5/1.15

Then you need for the numbers to have the same denominator.

After doing that the fraction would look like 114/39.9 for Quick Market and 173.48/39.9 for Stop and Save.

So Quick Market costs less.

3 0

4 years ago

The amount of warpage in a type of wafer used in the manufacture of integrated circuits has mean 1.3 mm and standard deviation 0

valina [46]

Answer:

a) $P(\bar X >1.305)=P(Z>\frac{1.305-1.3}{\frac{0.1}{\sqrt{200}}}=0.707)$

And using the complement rule, a calculator, excel or the normal standard table we have that:

$P(Z>0.707)=1-P(Z$

b) $z=-0.674$

And if we solve for a we got

$a=1.3 -0.674* \frac{0.1}{\sqrt{200}}=$

So the value of height that separates the bottom 95% of data from the top 5% is 1.295.

c) $P( \bar X >1.305) = 0.05$

We can use the z score formula:

$P( \bar X >1.305) = 1-P(\bar X$

Then we have this:

$P(z< \frac{1.305-1.3}{\frac{0.1}{\sqrt{n}}}) = 0.95$

And a value that accumulates 0.95 of the area on the normal distribution z = 1.64 and we can solve for n like this:

$n = (1.64*\frac{0.1}{1.305-1.3})^2= 1075.84 \approx 1076$

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

Let X the random variable that represent the amount of warpage of a population and we know

Where $\mu=1.3$ and $\sigma=0.1$

Since the sample size is large enough we can use the central limit theorem andwe know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

We can find the probability required like this:

$P(\bar X >1.305)=P(Z>\frac{1.305-1.3}{\frac{0.1}{\sqrt{200}}}=0.707)$

And using the complement rule, a calculator, excel or the normal standard table we have that:

$P(Z>0.707)=1-P(Z$

Part b

For this part we want to find a value a, such that we satisfy this condition:

$P(\bar X>a)=0.75$ (a)

$P(\bar X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-0.674$

And if we solve for a we got

$a=1.3 -0.674* \frac{0.1}{\sqrt{200}}=$

So the value of height that separates the bottom 95% of data from the top 5% is 1.295.

Part c

For this case we want this condition:

$P( \bar X >1.305) = 0.05$

We can use the z score formula:

$P( \bar X >1.305) = 1-P(\bar X$

Then we have this:

$P(z< \frac{1.305-1.3}{\frac{0.1}{\sqrt{n}}}) = 0.95$

And a value that accumulates 0.95 of the area on the normal distribution z = 1.64 and we can solve for n like this:

$n = (1.64*\frac{0.1}{1.305-1.3})^2= 1075.84 \approx 1076$

6 0

3 years ago

Peppy Pets charges a flat fee of $15 plus $3 per hour to keep a dog during the day. Happy Hounds charges a flat fee of $21 plus

stich3 [128]

Answer: d) 3 hours.

Step-by-step explanation:

Let x be the required number of hours for which the total fee charged by the companies the same.

Given: Peppy Pets charges a flat fee of $15 plus $3 per hour .

So Total charge = 15+3x ( in dollars)

Happy Hounds charges a flat fee of $21 plus $1 per hour.

So Total charge = 21+x ( in dollars)

When the total charge for both companies are the same, then

$15+3x=21+x\\\\\Rightarrow\ 3x-x=21-15\\\\\Rightarrow\ 2x=6\\\\\Rightarrow\ x=3$

Hence, the correct option is d) 3 hours.

7 0

3 years ago