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3 years ago

Work out 5 x 10^6 / 4 x 10^4 Give your answer as an ordinary number.

Mathematics

1 answer:

Angelina_Jolie [31]3 years ago

5 0

Answer:

$1.25 \times {10}^{2} = 125$

Step-by-step explanation:

$\frac{5 \times {10}^{6} }{4 \times {10}^{4} } = \frac{5}{4} \times \frac{ {10}^{6} }{ {10}^{4} } = 1.25 \times {10}^{2} = 125$

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tresset_1 [31]

The values on the horizontal axis are:
at 0 = 64
at one standard deviation (lower, upper) = (60.5 , 67.5)
at two standard deviation (lower, upper) = (57 , 71)
at three standard deviation (lower, upper) = (53.5 , 74,5)

B. P(x > 71) = 1 - P(x < 71) = 1 - P[z < (71 - 64)/3.5] = 1 - P(z < 2) = 1 - 0.97725 = 0.02275
Therefore the no of plants taller than 71 inches will be approximately 0.02275 * 3000 = 68

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3 years ago

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Answer:

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3 years ago

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Choose the equation of the line through (4, –5) and perpendicular to 6x + 12y = 13.

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Answer:

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Step-by-step explanation:

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3 years ago

Jaime has two fractio} , thns, 3/18 and 13/15 at he needs to compare. He knows he will not be able to tell which is bigger until

Mrrafil [7]

3/18 = 15/90
13/15 =78/90
To rewrite it as the same denominator you have to find a number that they both go into. So 90 is the lowest number that both 18 and 15 go into. 18 x 5 is 90 and 15 x 6 is 90 so what you do to the denominator you have to do to the numerator. So then you do 3 x 5 which equals 15 and you have to do 13 x 6 which equals 78. So the equation would end up being with the two fractions 15/90 and 78/90.

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3 years ago

Barron's reported that the average number of weeks an individual is unemployed is 18.5 weeks. Assume that for the population of

Ket [755]

Answer:

A) The sampling distribution for a sample size n=50 has a mean of 18.5 weeks and a standard deviation of 0.849.

B) P = 0.7616

C) P = 0.4441

Step-by-step explanation:

We assume that for the population of all unemployed individuals the population mean length of unemployment is 18.5 weeks and that the population standard deviation is 6 weeks.

A) We take a sample of size n=50.

The mean of the sampling distribution is equal to the population mean:

$\mu_s=\mu=18.5$

The standard deviation of the sampling distribution is:

$\sigma_s=\dfrac{\sigma}{\sqrt{n}}=\dfrac{6}{\sqrt{50}}=0.849$

B) We have to calculate the probability that the sampling distribution gives a value between one week from the mean. That is between 17.5 and 19.5 weeks.

We can calculate this with the z-scores:

$z_1=\dfrac{X_1-\mu}{\sigma/\sqrt{n}}=\dfrac{17.5-18.5}{6/\sqrt{50}}=\dfrac{-1}{0.8485}=-1.179\\\\\\z_2=\dfrac{X_2-\mu}{\sigma/\sqrt{n}}=\dfrac{19.5-18.5}{6/\sqrt{50}}=\dfrac{1}{0.8485}=1.179$

The probability it then:

$P(|X_s-\mu_s|$

C) For half a week (between 18 and 19 weeks), we recalculate the z-scores and the probabilities:

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{18-18.5}{6/\sqrt{50}}=\dfrac{-0.5}{0.8485}=-0.589$

$P(|X_s-\mu_s|$

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3 years ago