What is 7/10 - 1/4 (fraction)

The mk family ochard has 120 apple trees and 90 pear trees.if each fruit tree produces an average of 590 pounds of fruit per yea

denis-greek [22]

If there’s a total of 210 fruit trees, and each tree produces 590 pounds a year, you multiply both those numbers and get 123,900 pounds a year.

6 0

3 years ago

The distribution of cholesterol levels in teenage boys is approximately normal withnbsp mu equals 170 and sigma equals 30 (Sour

ddd [48]

Answer: 0.0336

Step-by-step explanation:

Given : The distribution of cholesterol levels in teenage boys is approximately normal with mean : $\mu= 170$

Standard deviation : $\sigma= 30$

The formula for z-score is given by :-

$z=\dfrac{x-\mu}{\sigma}$

For x=225

$z=\dfrac{225-170}{30}=1.83$

The p-value = $P(z>1.83)=1-P(z$

$=1-0.966375=0.033625\approx0.0336$

The probability that a teenage boy has a cholesterol level greater than 225 =0.0336

4 0

3 years ago

I need help with this question plz

Sauron [17]

Peter, to find the perimeter you have to add all sides up.

5 0

3 years ago

4. The average annual income of 100 randomly chosen residents of Santa Cruz is $30,755 with a standard deviation of $20,450. a)

Umnica [9.8K]

Answer:

a) The standard deviation of the annual income σₓ = 2045

b)

The calculated value Z = 0.608 < 1.645 at 10 % level of significance

Null hypothesis is accepted

The average annual income is greater than $32,000

c)

The calculated value Z = 1.0977 < 1.96 at 5 % level of significance

Null hypothesis is accepted

The average annual income is equal to $33,000

d)

95% of confidence intervals of the Average annual income

(26 ,746.8 ,34, 763.2)

Step-by-step explanation:

Given size of the sample 'n' =100

mean of the sample x⁻ = $30,755

The Standard deviation = $20,450

a)

The standard deviation of the annual income σₓ = $\frac{S.D}{\sqrt{n} }$

= $\frac{20,450}{\sqrt{100} }= 2045$

b)

Given mean of the Population μ = $32,000

Given size of the sample 'n' =100

mean of the sample x⁻ = $30,755

The Standard deviation ( σ)= $20,450

Null Hypothesis:- H₀: μ > $32,000

Alternative Hypothesis:H₁: μ < $32,000

Level of significance α = 0.10

$Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }$

$Z = \frac{30755-32000 }{\frac{20450}{\sqrt{100} } }$

Z= |-0.608| = 0.608

The calculated value Z = 0.608 < 1.645 at 10 % level of significance

Null hypothesis is accepted

The average annual income is greater than $32,000

c)

Given mean of the Population μ = $33,000

Given size of the sample 'n' =100

mean of the sample x⁻ = $30,755

The Standard deviation ( σ)= $20,450

Null Hypothesis:- H₀: μ = $33,000

Alternative Hypothesis:H₁: μ ≠ $33,000

Level of significance α = 0.05

$Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }$

$Z = \frac{30755-33000 }{\frac{20450}{\sqrt{100} } }$

Z = -1.0977

|Z|= |-1.0977| = 1.0977

The 95% of z -value = 1.96

The calculated value Z = 1.0977 < 1.96 at 5 % level of significance

Null hypothesis is accepted

The average annual income is equal to $33,000

d)

95% of confidence intervals is determined by

$(x^{-} - 1.96 \frac{S.D}{\sqrt{n} } , x^{-} + 1.96 \frac{S.D}{\sqrt{n} })$

$(30755 - 1.96 \frac{20450}{\sqrt{100} } , 30755 +1.96 \frac{20450}{\sqrt{100} })$

( 30 755 - 4008.2 , 30 755 +4008.2)

95% of confidence intervals of the Average annual income

(26 ,746.8 ,34, 763.2)

4 0

3 years ago

Please help me solve this problem. I'm stuck!

mezya [45]

Answer:

a. 4 hours

b. if studio b increases' its hourly price, the two studios will never have the same price.

Step-by-step explanation:

The function for studio 1 is

cost = 8(hours) + 10

studio 2

cost = 6(hours) + 16

they cost the same after 4 hours. making a table really helps.

8 0

3 years ago