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3 years ago

I need help on these questions ( will give brianliest if im able to )

Mathematics

1 answer:

kompoz [17]3 years ago

3 0

Answer:

Step-by-step explanation:

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The number 7 is what percent of 80

xz_007 [3.2K]

Answer:

5.6

Step-by-step explanation:because if you divide correctly it will equal 5.6

3 0

2 years ago

Find the value of x. RZ = 30 SW = 5x - 20

Murrr4er [49]

|RZ|=0.5|SW| therefore 2|RZ|=|SW|.

5x - 20 = 2 · 30

5x - 20 = 60 |add 20 to both sides

5x = 80 |divide both sides by 5

x = 16

5 0

3 years ago

Olve. Do your work on a separate she

Mazyrski [523]

Answer:

Step-by-step explanation:

2 coins are 0.25 = 50 cents. That leaves 92-50 = 42 cents and 5 coins to go.

The only way to get 2 cents is to have 2 pennies. That leaves 40 cents and 3 coins to go.

The only way to get 40 cents with 3 coins is 1 quarter, 1 dime, and 1 nickel.

So you have 3 quarters = 75 cents.

2 pennies = 2 cents.

1 dime = 10 cents.

1 nickel = 5 cents.

92 cents total and 7 coins.The only way to get 40 cents

5 0

3 years ago

Read 2 more answers

Need that help.......plz

Anika [276]

For Question Number one
(2+10)²+4=148

Question Number two:
(6+5)×(8-6)=22

5 0

3 years ago

Suppose that an airline overbooks seats on their flights. In particular, it sells 300 tickets for a flight when there are only 2

vladimir1956 [14]

Using the normal approximation to the binomial, it is found that there is a 0.994 = 99.4% probability that we will have enough seats for everyone who shows up.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.
The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

In this problem:

15% do not show up, so 100 - 15 = 85% show up, which means that $p = 0.85$ .
300 tickets are sold, hence $n = 300$ .

The mean and the standard deviation are given by:

$\mu = np = 300(0.85) = 255$

$\sigma = \sqrt{np(1-p)} = \sqrt{300(0.85)(0.15)} = 6.185$

The probability that we will have enough seats for everyone who shows up is the probability of at most 270 people showing up, which, using continuity correction, is $P(X \leq 270 + 0.5) = P(X \leq 270.5)$ , which is the p-value of Z when X = 270.5.

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

$Z = \frac{270.5 - 255}{6.185}$

$Z = 2.51$

$Z = 2.51$ has a p-value of 0.994.

0.994 = 99.4% probability that we will have enough seats for everyone who shows up.

A similar problem is given at brainly.com/question/24261244

8 0

3 years ago