Line j passes through points 1,6 and 7,1 line k is parallel to line j. What is the slope of line k?

The height, in inches, of a randomly chosen American woman is a normal random variable with mean μ = 64 and variance 2 = 7.84. (

Schach [20]

Answer:

Step-by-step explanation:

Given that the height in inches, of a randomly chosen American woman is a normal random variable with mean μ = 64 and variance 2 = 7.84.

X is N(64, 2.8)

Or Z = $\frac{x-64}{2.8}$

a) the probability that the height of a randomly chosen woman is between 59.8 and 68.2 inches.

$=P(59.8$

b) $P(X\geq 59)\\= P(X\geq -1.78)\\ \\=0.9625$

c) For 4 women to be height 260 inches is equivalent to

4x will be normal with mean (64*4) and std dev (2.8*4)

4x is N(266, 11.2)

$P(4x>260)= \\P(Z\geq -0.53571)\\=0.7054$

d) Z is N(0,1)

E(Z19) = $P(Z>19)\\= 0.000$

since normal distribution is maximum only between 3 std deviations form the mean on either side.

3 0

3 years ago

A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 419 gram setting. It is

Dahasolnce [82]

Answer:

The calculated t- value = 1.09 > 3.19 at 0.025 level of significance

Null hypothesis is rejected

There is sufficient evidence to support the claim that the bags are underfilled

Step-by-step explanation:

Step(i):-

Given the mean of the Population(μ) = 419 grams

The sample size 'n' =19

Given mean of the sample x⁻ = 412

Given variance of the sample S² = 784

S = √784 = 28

Step(ii):-

Null hypothesis: H₀: There is no sufficient evidence to support the claim that the bags are underfilled.

Alternative hypothesis: H₁:

There is sufficient evidence to support the claim that the bags are underfilled.

Test statistic

$t = \frac{x^{-} -mean}{\frac{S}{\sqrt{n} } }$

$t = \frac{412 -419}{\frac{28}{\sqrt{19} } }$

t = -1.09

|t| = |-1.09|

Degrees of freedom ν = n-1 = 19-1 =18

t₀.₀₂₅ = 3.19

Final answer:-

The calculated t- value = 1.09 > 3.19 at 0.025 level of significance

Null hypothesis is rejected

There is sufficient evidence to support the claim that the bags are underfilled

8 0

3 years ago

What if the carrots were 50 cents less per pound how much Victoria pay for 8.5 pounds of carrots?

nikdorinn [45]

If you multiply 8.5 by 50 cents your answer will be 4.25

5 0

3 years ago

There are four steps in the manufacturing process of a stuffed toy: cutting, stuffing, sealing, and packaging. There are two emp

Anit [1.1K]

Answer:

(a) 2min/15units

(b) 1min/20units

(c) The bottleneck stage is cutting. The process capacity is 450units/hour

(d) The capacity of the new employee should be 1unit/2sec. The process capacity will increase four times

Step-by-step explanation:

(a) Cycle time for stage cutting for 2 employees = 16sec/2units × 1min/60sec = 2min/15units

(b) Cycle time for stage sealing for 1 employee = 3sec/1unit × 1min/60sec = 1min/20units

(c) Process capacity of stage cutting = 15units/2min × 60min/1hour = 450units/hour

(d) If the capacity of the new employee is 1unit/2sec, the process capacity = 1unit/2sec × 3600sec/1hour = 1800units/hour. The process capacity increases 4 times (1800/450 = 4)

7 0

3 years ago

Find dy/dx if y =x^3+5x+2/x²-1

stiks02 [169]

Differentiate using the Quotient Rule –

$\qquad$ $\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\$

According to the given question, we have –

f(x) = x^3+5x+2
g(x) = x^2-1

Let's solve it!

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2) \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5) - ( x^3+5x+2) 2x}{(x^2-1)^2 }\\$

$\qquad$ $\pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\$

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\$

$\qquad$ $\pink{\therefore \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}} = \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\$

7 0

1 year ago