Boxes of granola are on sale at a price of 2 for $4.50. There are 12 ounces of granola on each box. What is the unit price in dollars per pound? (16 ounces = 1 pound)

Answers

Answer 1

Answer:

Answer:

3 dollars per pound

Step-by-step explanation:

First, you find the cost of a box of granola 4.5/2 = 2.25. Since there are 12 ounces of granola in each box, you could divide 2.25 by 12 to get the cost of granola per ounce(0.1875) and multiply it by 16 to get the cost of a pound (3).

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Ways to write 58.25?

Answers

Answer:

58 1/4, 233/4, 5825%

Step-by-step explanation:

Answer:

You can write as a fraction, a percentage, and in word form

Step-by-step explanation:

Which decimal number is equivalent to 7/10

Answers

Answer:

0.7

Step-by-step explanation:

Answer:

0.7

Step-by-step explanation:

7/10 can also be written as 7÷10

7÷10=0.7

Is this right? please correct me if I'm wrong.

Answers

Answer:

Yes it seems that it is right

At a large bank, account balances are normally distributed with a mean of $1,637.52 and a standard deviation of $623.16. What is the probability that a simple random sample of 400 accounts has a mean that exceeds $1,650?

Answers

Answer:

$P(\bar X >1650)=P(Z>(1650-1637.52)/((623.16)/(√(400)))=0.401)$

And we can use the complement rule and we got:

$P(Z>0.401) =1-P(Z<0.401) = 1-0.656= 0.344$

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".

Solution to the problem

Let X the random variable that represent the bank account balances of a population, and for this case we know the distribution for X is given by:

$X \sim N(1637.52,623.16)$

Where $\mu=1637.52$ and $\sigma=623.16$

Since the distribution of X is normal then the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, (\sigma)/(√(n)))$

And we can use the z score formula given by:

$z = (\bar X -\mu)/((\sigma)/(√(n)))$

And using this formula we got:

$P(\bar X >1650)=P(Z>(1650-1637.52)/((623.16)/(√(400)))=0.401)$

And we can use the complement rule and we got:

$P(Z>0.401) =1-P(Z<0.401) = 1-0.656= 0.344$

Answer: the probability is 0.49

Step-by-step explanation:

Since the account balances at the large bank are normally distributed.

we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = account balances.

µ = mean account balance.

σ = standard deviation

From the information given,

µ = $1,637.52

σ = $623.16

We want to find the probability that a simple random sample of 400 accounts has a mean that exceeds $1,650. It is expressed as

P(x > 1650) = 1 - P(x ≤ 1650)

For x = 1650,

z = (1650 - 1637.52)/623.16 = 0.02

Looking at the normal distribution table, the probability corresponding to the z score is 0.51

P(x > 1650) = 1 - 0.51 = 0.49

-3=?-11 Find the missing value.

Answers

Answer:

?=8

Step-by-step explanation:

I substituted the ? for x

-3=x-11

first we want to isolate the x to do that, you have to find a way to move -11 to the other side. Well, x+0=x right? So all we have to do is find a way to go from -11 to 0! In this case you add 11 to both sides. This is very important so don't forget.

-3=x-11

+11 +11

-3+11=8

x=8

If you check your work, this should also go back in perfectly.

-3=8-11

-3=-11+8

I hope this helps!

Have a good day!

Calculate|4+7π| . The absolute value of 4+7π is equal to the square root