A car traveled 180 miles at a constant rate.

Mathematics

1 answer:

ddd [48]3 years ago

7 0

I think answer should be the first one I hope this helps let me know if it’s correct or not okay thanks bye bye

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Factor the expression -45x^2 +5

klemol [59]

$-45x^2 +5 \\ \\ GCF = 5 \\ \\ 5( \frac{45x^2}{5} - \frac{5}{5} ) \\ \\ -5(9x^2-1) \\ \\ -5((3x)^2-1^2) \\ \\ -5(3x+1)(3x-1) \\ \\ Answer: \fbox {-5(3x+1)(3x-1)}$

5 0

3 years ago

Salaries for various positions can vary significantly depending on whether the company is in the public or private sector. The U

DIA [1.3K]

Answer:

a) 0.004

b) 0.9609

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 76503

Standard Deviation, σ = 8850

We are given that the distribution of annual salaries is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(salary greater than 100,000)

P(x > 100000)

$P( x > 100000) = P( z > \displaystyle\frac{100000 - 76503}{8850}) = P(z > 2.655)$

$= 1 - P(z \leq 2.655)$

Calculation the value from standard normal z table, we have,

$P(x > 100000) = 1 - 0.996 = 0.004 = 0.4\%$

b) P(sample mean annual salary falls between 70,000 and 80,000 dollars)

Sample size, n = 20

Standard error due to sampling =

$\dfrac{\sigma}{\sqrt{n}} = \dfrac{8850}{\sqrt{20}} = 1978.92$

$P(70000 \leq x \leq 80000) = P(\displaystyle\frac{70000 - 76503}{1978.92} \leq z \leq \displaystyle\frac{80000-76503}{1978.92})\\\\ = P(-3.286 \leq z \leq 1.767)\\\\= P(z \leq 1.767) - P(z < -3.286)\\=0.9614 - 0.0005= 0.9609 = 96.09\%$