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Charra [1.4K]
3 years ago
6

To make lemonade, you used one quart of water, one pint of lemon juice, and 9 ounces of your secret ingredient: honey. How much

lemonade did you make?
Mathematics
2 answers:
aleksley [76]3 years ago
8 0
1 quart= 32 ounces
1 pint= 16 ounces

32+16+9= 57 ounces

57 ounces= 7.125 cups

7.125 cups= 7 1/8 cups 
devlian [24]3 years ago
6 0
1 qt = 2 pt. 
2 pt. = 32 fl. oz. 

1 pt. = 16 fl. oz.

32 + 16 + 9= 57 oz. You make 57 fl. oz. of Lemonade. Hope I could help!. 
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In the figure shown below, if line l is parallel to line m, then find the value of x
Bezzdna [24]

Answer:

Step-by-step explanation:

x = 180-65 = 115°

7 0
3 years ago
A food-packaging apparatus underfills 10% of the containers. Find the probability that for any particular 10 containers the numb
Maksim231197 [3]

Answer:

a) P(X = 1) = 0.38742

b) P(X = 3) = 0.05740

c) P(X = 9) = 0.00000

d) P(X \geq 5) = 0.00163

Step-by-step explanation:

For each container, there are only two possible outcomes. Either it is undefilled, or it is not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem

There are 10 containers, so n = 10.

A food-packaging apparatus underfills 10% of the containers, so p = 0.1.

a) This is P(X = 1)

P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742

b) This is P(X = 3)

P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740

c) This is P(X = 9)

P(X = 9) = C_{10,9}.(0.1)^{9}.(0.9)^{1} = 0.00000

d) This is P(X \geq 5).

Either the number is lesser than five, or it is five or larger. The sum of the probabilities of each event is decimal 1. So:

P(X < 5) + P(X \geq 5) = 1

P(X \geq 5) = 1 - P(X < 5)

In which

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.34868

P(X = 1) = C_{10,1}.(0.1)^{1}.(0.9)^{9} = 0.38742

P(X = 2) = C_{10,2}.(0.1)^{2}.(0.9)^{8} = 0.1937

P(X = 3) = C_{10,3}.(0.1)^{3}.(0.9)^{7} = 0.05740

P(X = 4) = C_{10,4}.(0.1)^{1}.(0.9)^{9} = 0.38742

So

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.34868 + 0.38742 + 0.19371 + 0.05740 + 0.01116 = 0.99837

Finally

P(X \geq 5) = 1 - P(X < 5) = 1 - 0.99837 = 0.00163

3 0
3 years ago
If two of these shapes are randomly chosen, one after the other without replacement, what is the probability that the first will
aleksandrvk [35]

Answer:

P(T\ n\ S) = \frac{1}{14}

Step-by-step explanation:

*Missing Part of the Question*

4 stars

5 triangles

3 circles

3 squares

Required

Determine the probability of triangle being first then square being second

Total = 4 + 5 + 3 + 3

Total = 15

Represent the triangle with T and square with S

So, we're solving for P(T n S)

P(T\ n\ S) = P(T) * P(S)

Solving for P(T)

P(T) = \frac{n(T)}{Total}

P(T) = \frac{5}{15}

Solving for P(S)

The question implies a probability without replacement;

Hence Total has now been reduced by 1

Total = 14

P(S) = \frac{n(S)}{Total}

P(S) = \frac{3}{14}

Recall that

P(T\ n\ S) = P(T) * P(S)

P(T\ n\ S) = \frac{5}{15} * \frac{3}{14}

P(T\ n\ S) = \frac{15}{15 * 14}

P(T\ n\ S) = \frac{1}{14}

Hence, the required probability is

P(T\ n\ S) = \frac{1}{14}

3 0
3 years ago
Given m||n, find the value of x.<br> m<br> t<br> (5x-5)<br> (6x-27)
almond37 [142]

Step-by-step explanation:

due to the laws of symmetry of the angles of 2 crossing lines (and parallel lines are simply repeating and mirroring these angles), both angles must be equal.

5x - 5 = 6x - 27

-5 = x - 27

x = 22

4 0
2 years ago
32 53 24 54 61 13<br>17 36 04 10 51 68​
Leno4ka [110]

Answer:

<u>32 53 24 54 61 13</u>

<u>32 53 24 54 61 1317 36 04 10 51 68</u>

7 0
3 years ago
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