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3 years ago

Solve this equation, thank you!

Mathematics

2 answers:

melisa1 [442]3 years ago

8 0

Answer:

9.99

Step-by-step explanation:

uysha [10]3 years ago

6 0

Answer:

x = 9.99

Step-by-step explanation:

Using sine law , we have

Sine F/f = sine E/x

Sine 55deg/12 = Sine43deg/x

0.8192/12 = 0.6820/x

Cross multiply

0.8192 X x = 12 x 0.6820

0.8192x = 8.184

Divide both sides by 0.8192

0.8192x/0.8192=8.184/0.8192

x = 9.99

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Ms. Carr's Class Is Selling Magazines To Raise Money For A Field Trip. The Students In The Class Device They Wanted To Make 5.50

FinnZ [79.3K]

Here you are giving only the amount they want to raise (namely profit times number of magazines sold), and here you are also giving Money they want to raise... So clarifying, the money they want to raise, should include the money they will spend on buying the magazines (there is no statement saying they found them, or were given the magazines, so a cost should be involved)

Now if they are only making the count of "Field trip costs X amount of money, and given we have to make a profit of $5.5, How many must we sell?" then the equation should be n=X/5.5

Should the story be, how much money must they raise to have a profit of 5.5 on each magazine and still have enough for the field trip, then you have a different equation which varies only in adding the cost of each magazine, either case, M should be defined not as money they need to raise (cause here they will be short on their goal) but Money they must earn. And again, you should rewrite your equation to be:

M=Amount they must raise
C=Cost per magazine
n=Number of magazines
p=profit $5.5 per magazine

C+p=M/n

And rewriting the previous they should make:

n(C+p)=M -----> n(C+5.5)=M m/n = 5.50 
m/n x n = 5.50 x n //// multiply each side by n 

m = 5.5n

6 0

3 years ago

Read 2 more answers

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skelet666 [1.2K]

I’m sorry I don’t know

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3 years ago

The Petri dish shown has no lid. What is the surface area of the outside of the Petri dish?

Nata [24]

Answer: (not enough information) Is there a picture that goes along with the problem?

Step-by-step explanation:

3 0

3 years ago

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. Consider the sample m

love history [14]

Answer:

a) The expected value of the sample mean weight is 20.4 pounds.

b)The standard deviation of the sample mean weight is 0.123.

c) There is a 14.46% probability the sample mean weight will be less than 20.27.

d) This value is $c = 20.6153$ .

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be 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.

In this problem, we have that:

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. This means that $\mu = 20.4, \sigma = 1.23$

Consider the sample mean weight of 100 watermelons of this variety. This means that $n = 100$ .

a. What is the expected value of the sample mean weight? Give an exact answer.

By the Central Limit Theorem, it is the same as the mean of the population. So the expected value of the sample mean weight is 20.4 pounds.

b. What is the standard deviation of the sample mean weight? Give your answer to four decimal places.

By the Central Limit Theorem, that is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{1.23}{\sqrt{100}} = 0.123$

The standard deviation of the sample mean weight is 0.123.

c. What is the approximate probability the sample mean weight will be less than 20.27?

This is the pvalue of Z when $X = 20.27$ .

Since we are working with the sample mean, we use $s$ instead of $\sigma$ in the Z score formula

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

$Z = \frac{20.27 - 20.4}{0.123}$

$Z = -1.06$

$Z = -1.06$ has a pvalue of 0.1446.

This means that there is a 14.46% probability the sample mean weight will be less than 20.27.

d. What is the value c such that the approximate probability the sample mean will be less than c is 0.96?

This is the value of $X = c$ that is in the 96th percentile, that is, it's Z score has a pvalue 0.96.

So we use $Z = 1.75$

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

$1.75 = \frac{c - 20.4}{0.123}$

$c - 20.4 = 0.123*1.75$

$c = 20.6153$

This value is $c = 20.6153$ .

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3 years ago

Kevin needs 1,000 ribbons for a school event. Each ribbon has to be 2.25 feet long. Tell how Kevin can use a pattern to find how

marysya [2.9K]

2.25 x 1000 = 2250.

That's how much yds of ribbons he needs.

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3 years ago