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Arlecino [84]
3 years ago
12

Bob and Maria each purchase steak from the butcher. Maria selects a cut that costs $19.98 per pound. Bob’s selection costs $13.9

8 per pound. If Maria’s steak is all meat, and Bob’s cut of beef is 1⁄3 bone and fat, who is paying more for actual steak?
Mathematics
1 answer:
Sidana [21]3 years ago
8 0

Answer:

Bob is paying more for the actual steak

Step-by-step explanation:

we know that

Maria selects a cut that costs $19.98 per pound (Maria's steak is all meat)

Bob’s selection costs $13.98 per pound (Bob’s cut of beef is 1⁄3 bone and fat)

so

Bob's cut is 2/3 all meat

Find the unit rate of Bob's cut for all meat

\frac{13.98}{(2/3)} \frac{\$}{pounds} =20.97\frac{\$}{pounds}

therefore

\$20.97> \$19.98

Bob is paying more for the actual steak

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ratelena [41]
Any answer things? :|
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3 years ago
11. Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is F(x) 5 5 0 x , 0
NISA [10]

Question not properly presented

Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is F(x)

0 ------ x<0

x²/25 ---- 0 ≤ x ≤ 5

1 ----- 5 ≤ x

Use the cdf to obtain the following.

(a) Calculate P(X ≤ 4).

(b) Calculate P(3.5 ≤ X ≤ 4).

(c) Calculate P(X > 4.5)

(d) What is the median checkout duration, μ?

e. Obtain the density function f (x).

f. Calculate E(X).

Answer:

a. P(X ≤ 4) = 16/25

b. P(3.5 ≤ X ≤ 4) = 3.75/25

c. P(4.5 ≤ X ≤ 5) = 4.75/25

d. μ = 3.5

e. f(x) = 2x/25 for 0≤x≤2/5

f. E(x) = 16/9375

Step-by-step explanation:

a. Calculate P(X ≤ 4).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(X ≤ 4) = F(x) {0,4}

P(X ≤ 4) = x²/25 {0,4}

P(X ≤ 4) = 4²/25

P(X ≤ 4) = 16/25

b. Calculate P(3.5 ≤ X ≤ 4).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(3.5 ≤ X ≤ 4) = F(x) {3.5,4}

P(3.5 ≤ X ≤ 4) = x²/25 {3.5,4}

P(3.5 ≤ X ≤ 4) = 4²/25 - 3.5²/25

P(3.5 ≤ X ≤ 4) = 16/25 - 12.25/25

P(3.5 ≤ X ≤ 4) = 3.75/25

(c) Calculate P(X > 4.5).

Given that the cdf, F(x) = x²/25 for 0 ≤ x ≤ 5

So, we have

P(4.5 ≤ X ≤ 5) = F(x) {4.5,5}

P(4.5 ≤ X ≤ 5) = x²/25 {4.5,5}

P(4.5 ≤ X ≤ 5)) = 5²/25 - 4.5²/25

P(4.5 ≤ X ≤ 5) = 25/25 - 20.25/25

P(4.5 ≤ X ≤ 5) = 4.75/25

(d) What is the median checkout duration, μ?

Median is calculated as follows;

∫f(x) dx {-∝,μ} = ½

This implies

F(x) {-∝,μ} = ½

where F(x) = x²/25 for 0 ≤ x ≤ 5

F(x) {-∝,μ} = ½ becomes

x²/25 {0,μ} = ½

μ² = ½ * 25

μ² = 12.5

μ = √12.5

μ = 3.5

e. Calculating density function f (x).

If F(x) = ∫f(x) dx

Then f(x) = d/dx (F(x))

where F(x) = x²/25 for 0 ≤ x ≤ 5

f(x) = d/dx(x²/25)

f(x) = 2x/25

When

F(x) = 0, f(x) = 2(0)/25 = 0

When

F(x) = 5, f(x) = 2(5)/25 = 2/5

f(x) = 2x/25 for 0≤x≤2/5

f. Calculating E(X).

E(x) = ∫xf(x) dx, 0,2/5

E(x) = ∫x * 2x/25 dx, 0,2/5

E(x) = 2∫x ²/25 dx, 0,2/5

E(x) = 2x³/75 , 0,2/5

E(x) = 2(2/5)³/75

E(x) = 16/9375

4 0
3 years ago
Consider the following operations on the number 8.82 x 10^-2. Without using a calculator, decide which would give a significantl
Ksenya-84 [330]

Correct question is;

Consider the following operations on the number 8.85 x 10^(-2). Without using a calculator, decide which would give a significantly smaller value than 8.85 x 10^(-2), which would give a significantly larger value, or which would give essentially the same value.

A) 8.85 × 10^(-2) + 6.69 × 10^(5)

B) 8.85 x 10^(-2) - 6.69 x 10^(5)

C) 8.85 x 10^(-2) x 6.69 x 10^(5)

D) 8.85 × 10^(-2) ÷ 6.69 × 10^(5)

Answer:

Option A & B yield essentially the same value.

Option C yields a significantly smaller value

Option D yields a significantly larger value

Step-by-step explanation:

Since given value for the operations is in 10^(-2) indices, then let's convert the operations numbers from 10^(5) indices to 10^(-2).

Thus;

6.69 × 10^(5) = (6.69 × 10^(-7)) × 10^(-2)

Thus;

A) 8.85 × 10^(-2) + 6.69 × 10^(5) is now;

8.85 × 10^(-2) + ((6.69 × 10^(-7)) × 10^(-2))

This gives;

[8.85 + (6.69 × 10^(-7))] × 10^(-2)

The 10^(-7) indices indicates that the initial value of 8.85 × 10^(-2) would not change much and would essentially be the same

B) 8.85 x 10^(-2) - 6.69 x 10^(5) is now;

8.85 × 10^(-2) - ((6.69 × 10^(-7)) × 10^(-2))

This gives;

[8.85 - (6.69 × 10^(-7))] × 10^(-2)

Again the 10^(-7) indices indicates that the initial value of 8.85 × 10^(-2) would not change much and would essentially be the same.

C) 8.85 x 10^(-2) x 6.69 x 10^(5) is now;

(8.85 × 10^(-2)) × ((6.69 × 10^(-7)) × 10^(-2))

This gives;

[8.85 × (6.69 × 10^(-7))] × 10^(-2)

When we multiply 8.85 by 6.69 × 10^(-7), the value gotten will be way lower than the initial value of 8.85.

Thus,this operation will give a significantly smaller value.

D) (8.85 × 10-2) ÷ (6.69 × 10^5) is now;

(8.85 × 10^(-2)) ÷ ((6.69 × 10^(-7)) × 10^(-2))

This gives;

[8.85 ÷ (6.69 × 10^(-7))] × 10^(-2)

When we carry out the operation:

8.85 ÷ (6.69 × 10^(-7)), it is essentially equal to (8.85 × 10^(7))/6.69.

This means that the new value would be significantly larger than the initial value of 8.85.

Thus,this operation yields a significantly larger value.

4 0
3 years ago
HEY I NEED HELP (sorry in advance for my terrible handwriting)
chubhunter [2.5K]
< This is the answer!!!!!!
4 0
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Read 2 more answers
A point on the rim of a wheel has a linear speed of 33 cm/s. If the radius of the wheel is 50 cm, what is the angular speed of t
Zigmanuir [339]

Answer:

The angular speed of the wheel in radians per second is 0.66.

Step-by-step explanation:

Recall the following statement:

A linear speed (v) is given by,  

v=\omega \times r                       ...... (1)

Here,  \omega represents the angular speed of the wheel and <em>r</em> represents the  radius of the wheel.

From the given information:

Linear speed (v) = 33 cm/s

Radius of the wheel (r) = 50 cm

Now to find the angular speed in radian per second.

33 =50 \times \omega

Divide both sides by 50.

0.66 rad/sec= \omega

Hence, the angular speed of the wheel in radians per second is 0.66.

4 0
3 years ago
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