Which expression could be used to determine the product of -4 and 3 1/4

Recent studies indicate that the typical 50-year-old woman spends $350 per year for personal-care products. The distribution of

jek_recluse [69]

Answer:

P ( x_bar > 335 ) = 0.9826

Step-by-step explanation:

Given:

- Mean amount u = 350

- standard deviation s.d = 45/year

- Sample size n = 40

Find:

- The probability of sample mean P( x_bar > 335 )

Solution:

- P ( x_bar > 335 ) = P ( Z > sqrt(n)*(x_bar - u)/s.d)

= P ( Z > sqrt(40)*(335-350)/45)

= P ( Z > -2.111) = P ( Z < 2.111)

= 0.5 + P( 0 < Z < 2.111)

= 0.5 + 0.4826

= 0.9826

4 0

1 year ago

Match each pair of points to the equation of the line that is parallel to the line passing through the points.

Paraphin [41]

we know that

If two lines are parallel, then, their slopes are equal.

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we will proceed to calculate the slope in each case, to determine the solution of the problem

Case A) Point $B(5,2)\ C(7,-5)$

Find the slope BC

Substitute the values in the formula

$m=\frac{-5-2}{7-5}$

$m=\frac{-7}{2}$

$m=-3.5$

so

The equation $y=-3.5x-15$ is parallel to the line passing through the points $B(5,2)\ C(7,-5)$

therefore

the answer Part A) is

$B(5,2)\ C(7,-5)$ ------> $y=-3.5x-15$

Case B) Point $D(11,6)\ E(5,9)$

Find the slope DE

Substitute the values in the formula

$m=\frac{9-6}{5-11}$

$m=\frac{3}{-6}$

$m=-0.5$

so

The equation $y=-0.5x-3$ is parallel to the line passing through the points $D(11,6)\ E(5,9)$

therefore

the answer Part B) is

$D(11,6)\ E(5,9)$ ------> $y=-0.5x-3$

Case C) Point $F(-7,12)\ G(3,-8)$

Find the slope FG

Substitute the values in the formula

$m=\frac{-8-12}{3+7}$

$m=\frac{-20}{10}$

$m=-2$

so

Any linear equation with slope $m=-2$ will be parallel to the line passing through the points $F(-7,12)\ G(3,-8)$

Case D) Point $H(4,4)\ I(8,9)$

Find the slope HI

Substitute the values in the formula

$m=\frac{9-4}{8-4}$

$m=\frac{5}{4}$

$m=1.25$

so

The equation $y=1.25x+4$ is parallel to the line passing through the points $H(4,4)\ I(8,9)$

therefore

the answer Part D) is

$H(4,4)\ I(8,9)$ ------> $y=1.25x+4$

Case E) Point $J(7,2)\ K(-9,8)$

Find the slope JK

Substitute the values in the formula

$m=\frac{8-2}{-9-7}$

$m=\frac{6}{-16}$

$m=-0.375$

so

Any linear equation with slope $m=-0.375$ will be parallel to the line passing through the points $J(7,2)\ K(-9,8)$

Case F) Point $L(5,-7)\ M(4,-12)$

Find the slope LM

Substitute the values in the formula

$m=\frac{-12+7}{4-5}$

$m=\frac{-5}{-1}$

$m=5$

so

The equation $y=5x+19$ is parallel to the line passing through the points $L(5,-7)\ M(4,-12)$

therefore

the answer Part F) is

$L(5,-7)\ M(4,-12)$ ------> $y=5x+19$

8 0

1 year ago

Read 2 more answers

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard dev

Nataliya [291]

Answer:

(a) Probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long = 0.15716

(b) Probability that a standard normal random variable will be between .3 and 3.2 = 0.3814

Step-by-step explanation:

We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with;

Mean, $\mu$ = 30.05 inches and Standard deviation, $\sigma$ = 0.2 inches

Let X = A sheet selected at random from the population

Here, the standard normal formula is ;

Z = $\frac{X - \mu}{\sigma}$ ~ N(0,1)

(a) The Probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long = P(30.25 < X < 30.65)

P(30.25 < X < 30.65) = P(X < 30.65) - P(X <= 30.25)

P(X < 30.65) = P( $\frac{X - \mu}{\sigma}$ < $\frac{30.65 - 30.05}{0.2}$ ) = P(Z < 3) = 1 - P(Z >= 3) = 1 - 0.001425

= 0.9985

P(X <= 30.25) = P( $\frac{X - \mu}{\sigma}$ <= $\frac{30.25 - 30.05}{0.2}$ ) = P(Z <= 1) = 0.84134

Therefore, P(30.25 < X < 30.65) = 0.9985 - 0.84134 = 0.15716 .

(b) Let Y = Standard Normal Variable is given by N(0,1)

Which means mean of Y = 0 and standard deviation of Y = 1

So, Probability that a standard normal random variable will be between 0.3 and 3.2 = P(0.3 < Y < 3.2) = P(Y < 3.2) - P(Y <= 0.3)

P(Y < 3.2) = P( $\frac{Y - \mu}{\sigma}$ < $\frac{3.2 - 0}{1}$ ) = P(Z < 3.2) = 1 - P(Z >= 3.2) = 1 - 0.000688

= 0.99931

P(Y <= 0.3) = P( $\frac{Y - \mu}{\sigma}$ <= $\frac{0.3 - 0}{1}$ ) = P(Z <= 0.3) = 0.61791

Therefore, P(0.3 < Y < 3.2) = 0.99931 - 0.61791 = 0.3814 .

3 0

2 years ago

297,060 / 0.0004839 = Answer in Scientific Notation it Standard Notation

Valentin [98]

Answer:

6.1389 x 10^8

Step-by-step explanation:

3 0

1 year ago

Write 55,426 using the number names

Aleks04 [339]

Answer:

fifty five thousand four hundred and twenty six.

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers