(50 points) URGENT <br> Can someone help me with this problem?

HELP PLEASE!!!

amm1812

$2300\cdot\frac{6}{100}=23*6=138$ ears of non edible corn

6 0

3 years ago

There are 5 children and 3 adults going on a trip.

sergij07 [2.7K]

Answer:

3/8

Step-by-step explanation:

8 0

3 years ago

The diameter of a particle of contamination (in micrometers) is modeled with the probability density function f(x)= 2/x^3 for x

RoseWind [281]

Answer:

a) 0.96

b) 0.016

c) 0.018

d) 0.982

e) x = 2

Step-by-step explanation:

We are given with the Probability density function f(x)= 2/x^3 where x > 1.

<em>Firstly we will calculate the general probability that of P(a < X < b) </em>

P(a < X < b) = $\int_{a}^{b} \frac{2}{x^{3}} dx$ = $2\int_{a}^{b} x^{-3} dx$

= $2[ \frac{x^{-3+1} }{-3+1}]^{b}_a dx$ { Because $\int_{a}^{b} x^{n} dx = [ \frac{x^{n+1} }{n+1}]^{b}_a$ }

= $2[ \frac{x^{-2} }{-2}]^{b}_a$ = $\frac{2}{-2} [ x^{-2} ]^{b}_a$

= $-1 [ b^{-2} - a^{-2} ]$ = $\frac{1}{a^{2} } - \frac{1}{b^{2} }$

a) Now P(X < 5) = P(1 < X < 5) {because x > 1 }

Comparing with general probability we get,

P(1 < X < 5) = $\frac{1}{1^{2} } - \frac{1}{5^{2} }$ = $1 - \frac{1}{25}$ = 0.96 .

b) P(X > 8) = P(8 < X < ∞) = 1/ $8^{2}$ - 1/∞ = 1/64 - 0 = 0.016

c) P(6 < X < 10) = $\frac{1}{6^{2} } - \frac{1}{10^{2} }$ = $\frac{1}{36} - \frac{1}{100 }$ = 0.018 .

d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)

= $(\frac{1}{1^{2} } - \frac{1}{6^{2} })$ + (1/ $10^{2}$ - 1/∞) = 1 - 1/36 + 1/100 + 0 = 0.982

e) We have to find x such that P(X < x) = 0.75 ;

⇒ P(1 < X < x) = 0.75

⇒ $\frac{1}{1^{2} } - \frac{1}{x^{2} }$ = 0.75

⇒ $\frac{1} {x^{2} }$ = 1 - 0.75 = 0.25

⇒ $x^{2}$ = $\frac{1}{0.25}$ ⇒ $x^{2}$ = 4 ⇒ x = $2$

Therefore, value of x such that P(X < x) = 0.75 is 2.

8 0

3 years ago

9(k+12)+4k2 when k= 3 the 2 is small

Advocard [28]

Answer:

the answer is 4k^2+9k+108

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

A study was performed to investigate whether teens and adults had different habits when it comes to consuming meat-free meals. I

Elden [556K]

Answer:

H₀ should be rejected at CI 95% .

Then the proportion of adults with 95 % CI is bigger than the proportion of teen

Step-by-step explanation:

From adult sample:

n₂ = 2323

x₂ = 1601

p₂ = 1601 / 2323 p₂ = 0,689 or p₂ = 68,9%

From teen sample:

n₁ = 875

x₁ = 555

p₁ = 555/ 875 p₁ = 0,634 or p₁ = 63,4

Values of p₁ and p₂ suggest that the proportion of adults consuming at least one meat free meal per week is bigger than teen proportion.

To either prove or reject the above statement we have to develop a difference of proportion test according to:

Hypothesis test:

Null Hypothesis H₀ p₁ = p₂

Alternative Hypothesis Hₐ p₂ > p₁

So is a one-tail test to the right

We can establish a confidence interval of 95 % then α = 5 %

or α = 0,05

As the samples are big enough we will develop a z test

Then z(c) for α = 0,05 from z table is z(c) = 1,64

To calculate z(s)

z(s) = ( p₂ - p₁ ) / √p*q* ( 1/n₁ + 1/n₂ )

where p = ( x₁ + x₂ ) / n₁ + n₂ p = 555 + 1601 / 875 + 2323

p = 2156/3198 p = 0,674

and q = 1 - 0,674 q = 0,326

z(s) = ( 0,689 - 0,634 ) / √0,674*0,326 ( 1/875 + 1 / 2323

z(s) = 0,055/ √ 0,2197 ( 0,00114 + 0,00043)

z(s) = 0,055/ √0,2197* 0,00157

z(s) = 0,055/ √ 3,45*10⁻⁴

z(s) = 0,055 / 1,85*10⁻²

z(s) = 5,5/1,85

z(s) = 2,97

Comparing z(s) and z(c) z(s) > z(c)

Then z(s) is in the rejection region we reject H₀.

We can claim that the proportion of adult eating at least one meat-free meal is bigger than the proportion of teen

6 0

3 years ago

(50 points) URGENT Can someone help me with this problem?