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tigry1 [53]
3 years ago
14

Paco says that the solution to 2/5x - 1 = 9 is x = 4. Do you agree? Explain.

Mathematics
1 answer:
QveST [7]3 years ago
8 0

Answer:

yes x=4

Step-by-step explanation:

2/5x -1=9

+1 +1

2(2/5x)=(10)2

5x/5=20/5

X=4

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Solve the equation 6(y+1.5)= -18. what is the value of y? y=
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y = -4.5

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Para una función de teatro para niños, el boleto para adultos cuesta $70 y el boleto para niños $35. Anoche entraron en total 80
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frez [133]

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7 0
3 years ago
Can someone help me with this!! I’ve been stuck for awhile!!
scoundrel [369]

Answer:

(C) h = \frac{2a}{b}

Step-by-step explanation:

Here we have the formula for the area of a triangle: A = \frac{1}{2}bh. Our goal is to isolate h on one side of the equation.

To do this, let’s treat the rest of the equation as if there were no variables.

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2a = bh

Divide both sides by b:

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Hope this helped!

3 0
3 years ago
A scientist claims that 9% of viruses are airborne. If the scientist is accurate, what is the probability that the proportion of
Lana71 [14]

The probability that the sample proportion of airborne viruses in differ from the population proportion by greater than 3% is 0.006.

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

μρ=ρ

The standard deviation of this sampling distribution of sample proportion is:

σρ=\sqrt \frac{p(1-p)}{n}

The information provided is:

n = 533

p = 0.09

As the sample size is quite large, i.e. n = 533 > 30, the central limit theorem can be used to approximate the sampling distribution of sample proportion by a Normal distribution.

The mean and standard deviation are:

μρ=ρ=0.09

σρ=\sqrt \frac{p(1-p)}{n}=\sqrt\frac{0.09(1-0.09)}{533}=0.01239

Compute the probability that the sample proportion of airborne viruses in differ from the population proportion by greater than 3% as follows:

p(p^-p > 0.03)=P(p^ > 0.12)                     \\=P(z > \frac{0.12-0.9}{0.012})                      \\= P(Z > 2.5)                     \\= 1-P(Z > 2.5)                    \\ = 1- 0.993                   \\  = 0.00621

Thus, the probability that the sample proportion of airborne viruses in differ from the population proportion by greater than 3% is 0.006.

Learn more about PROBABILITY here

brainly.com/question/24756209

#SPJ4

4 0
2 years ago
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