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3 years ago

There are 10 girls and 8 boys in Mr Augello's class

Mathematics

2 answers:

poizon [28]3 years ago

7 0

Whats the rest of the question

BARSIC [14]3 years ago

5 0

that would be 18 ppl in his class not counting him self

i think thayts what your trying to ask hope this helps

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Find the solution to the system of equations.

harina [27]

Answer:

x=2

y=3

Step-by-step explanation:

y=−2x+7

y=5x−7

Set the two equations equal since they are both equal to y

−2x+7 =5x−7

Add 2x to each side

-2x+7+2x = 5x-7+2x

7 = 7x-7

Add 7 to each side

7+7 = 7x-7+7

14 =7x

Divide by 7

14/7 = 7x/7

2 =x

Now find 7

y = 5x-7

y = 5(2) -7

y = 10-7

y = 3

4 0

3 years ago

Read 2 more answers

Cora gave her a $4.40 tip. The tip was 20% of the cost of the haircut. Write an equation to find b, the cost of the haircut .

Damm [24]

Answer:

3.52

Step-by-step explanation:

do 4.40*.20. youll get .88, then subtract .88 from 4.40. pls give brainliest

7 0

2 years ago

If <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bdy%7D%7Bdt%7D%20%3Dky" id="TexFormula1" title=" \frac{dy}{dt} =ky" alt=" \frac

sladkih [1.3K]

This ODE is separable; we have

$\dfrac{\mathrm dy}{\mathrm dt}=ky\implies\dfrac{\mathrm dy}y=k\,\mathrm dt$

Integrating both sides gives a general solution of

$\ln|y|=kt+C\implies y=e^{kt+C}=Ce^{kt}$

B is the only choice that is applicable.

4 0

3 years ago

The scores of students on the ACT college entrance exam in a recent year had the normal distribution with mean  =18.6 and stand

Maurinko [17]

Answer:

a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) 0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

Step-by-step explanation:

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

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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 18.6, \sigma = 5.9$

a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?

This is 1 subtracted by the pvalue of Z when X = 21. So

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

$Z = \frac{21 - 18.6}{5.4}$

$Z = 0.44$

$Z = 0.44$ has a pvalue of 0.67

1 - 0.67 = 0.33

33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have $n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768$