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3 years ago

Combine like terms. 3y-4-6-3x-7y+6x+1

Mathematics

1 answer:

maks197457 [2]3 years ago

4 0

Answer:

If you combine everything its -4y+3x-9

Step-by-step explanation:

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Use order of operation and the mathematical properties of numbers to simplify these numbers (3+2^2)^2 =

Art [367]

Answer:

Step-by-step explanation:

(3+2²)²

(3+4)²

(7)²

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3 years ago

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The price of milk has been increasing over the last month. Audrey believes there is a positive correlation between the number of

iris [78.8K]

Answer:

Its 0.33

Step-by-step explanation:

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4 0

3 years ago

Find p such that p+1=2x^2, p^2+1=2y^2. Know that p is a prime, x and y are positive integers.

Charra [1.4K]

Answer:

Step-by-step explanation:

Given p + 1 = 2 $x^{2}$ → 1 = 2 $x^{2}$ - p

$p^{2} + 1 = 2y^{2}$

$p^{2} + (2x^{2} - p) = 2y^{2}$

$p^{2} - p + (2x^{2} - 2y^{2} ) = 0$

The above equation is quadratic in p.

p = [ -(-1) ± $\sqrt{(-1)^{2} - 4* 1*(2x^{2} - 2y^{2} )$ ] ÷ 2

p = [1 ± $\sqrt{1 - 8(x^{2}-y^{2} )}$ ] ÷ 2

4 0

3 years ago

Noa buys 6 items that each cost $4.25 and 3 items that each cost $6.75. How much does Noa spend in total?

den301095 [7]

Answer:

Answer is D - $45.75

Step-by-step explanation:

(6 x $4.25) + (3 x $6.75) = $45.75

4 0

3 years ago

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A scientist claims that 9% of viruses are airborne.

Aleksandr [31]

Answer:

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

Step-by-step explanation:

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:

$\mu_{\hat p}=p$

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

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}$

The information provided is:

n = 596

p = 0.09

As the sample size is quite large, i.e. n = 596 > 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:

$\mu_{\hat p}=p=0.09\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.09(1-0.09)}{596}}=0.012$

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

$P(\hat p-p>0.03)=P(\hat p>0.12)$

$=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}>\frac{0.12-0.09}{0.012})\\\\=P(Z>2.5)\\\\=1-P(Z$

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

5 0

3 years ago