Please help with this

A company produces fruity drinks that contain a percentage of real fruit juice. Drink A contains 5% real fruit juice and Drink B

Romashka [77]

Answer:

a). 12.5 liters of real fruit juice would be needed to produce 250 liters of Drink A, and 20 liters of real fruit juice would be needed to produce 250 liters of Drink B

b). 0.05aa liters of real fruit juice would be needed to produce aa liters of Drink A, and 0.1bb liters of real fruit juice would be needed to produce bb liters of Drink B

Step-by-step explanation:

a). Drinks A(250 liters) and Drink B(200 liters)

Total amount of Drink A=250 liters

Real fruit juice=5% of Drink A=(5/100)×250=12.5 liters

12.5 liters of real fruit juice would be needed to produce 250 liters of Drink A

Total amount of Drink B=200 liters

Real fruit juice=10% of Drink B=(10/100)×200=20 liters

20 liters of real fruit juice would be needed to produce 250 liters of Drink B

b). Drinks aa and bb

Total amount of Drink A=aa liters

Real fruit juice=5% of Drink A=(5/100)×aa=0.05aa liters

0.05aa liters of real fruit juice would be needed to produce aa liters of Drink A

Total amount of Drink B =bb liters

Real fruit juice=10% of Drink 10=(10/100)×bb=0.1bb liters

0.1bb liters of real fruit juice would be needed to produce bb liters of Drink B

5 0

4 years ago

Someone Help I will give Branlist for whoever answers first!!!

Assoli18 [71]

Answer:

A its like, getting a plate before you make a sandwich, you have to get a plate first, or else you can't start making your sandwich.

Step-by-step explanation:

you don't have to give brainliest, im just helping lol

6 0

2 years ago

Read 2 more answers

Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probabili

lawyer [7]

Answer:

(i) The probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

Step-by-step explanation:

Denote the events as follows:

B = bugs are present in a computer program.

D = a de-bugging program detects the bug.

The information provided is:

$P(B) =0.01\\P(D|B)=0.99\\P(D|B^{c})=0.02$

(i)

The probability that there is a bug in the program given that the de-bugging program has detected the bug is, P (B | D).

The Bayes' theorem states that the conditional probability of an event E given that another event X has already occurred is:

$P(E|X)=\frac{P(X|E)P(E)}{P(X|E)P(E)+P(X|E^{c})P(E^{c})}$

Use the Bayes' theorem to compute the value of P (B | D) as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|B^{c})P(B^{c})}=\frac{(0.99\times 0.01)}{(0.99\times 0.01)+(0.02\times (1-0.01))}=0.3333$

Thus, the probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii)

The probability that a bug is actually present given that the de-bugging program claims that bug is present is:

P (B|D) = 0.3333

Now it is provided that two tests are performed on the program A.

Both the test are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is:

P (Bugs are actually present | Detects on both test) = P (B|D) × P (B|D)

$=0.3333\times 0.3333\\=0.11108889\\\approx 0.1111$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii)

Now it is provided that three tests are performed on the program A.

All the three tests are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is:

P (Bugs are actually present | Detects on all 3 test)

= P (B|D) × P (B|D) × P (B|D)

$=0.3333\times 0.3333\times 0.3333\\=0.037025927037\\\approx 0.037$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

4 0

3 years ago

You just reflected on working with right triangles and trigonometric concepts. How important were concepts that you learned prev

harina [27]

Idk if ya still need it or if this will help but this is what i put for mine

Concepts that I had previously learned about are quite important because math goes hand and hand. if you try to skip something and go ahead you'll most likely get confused because there could of been something highly important to that would help you understand whats ahead. in the previous unit it tells you about triangles and their description like a right triangle-90 degree angle and where legs and hypotenuse is located, this unit has had a ton about right triangles.

4 0

3 years ago

GR5_Math_U7_FS

Nutka1998 [239]

Answer:

4/6 of a pizza.

Step-by-step explanation:

We know that:

Both pizzas are of the same size.

Jasmine cut her's into 12 slices.

Kelly cut her's into 6 slices, and ate 2 slices, then Kelly ate 2/6 of a pizza.

Jasmine ate X of a pizza, and we know that between them ate a total of 1 pizza.

This means that:

X of a pizza + 2/6 of a pizza = 1 pizza

X + 2/6 = 1

X = 1 - 2/6

X = 6/6 - 2/6 = 4/6

Jasmine ate 4/6 of a pizza, knowing that she cut her's into 12 slices, this would mean that she ate 8/12 of a pizza, or 8 slices.

5 0

3 years ago