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Ghella [55]
3 years ago
12

How do I find the intervals when it's set up as a fraction?

Mathematics
1 answer:
Hoochie [10]3 years ago
3 0
You divide by the 2x by 5 and thats your answer :)

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The Cunningham family went to a dinner at Pasta Palace. Mr Cunningham ordered a meal for &7.50; Mrs. C ordered a meal for $6
SIZIF [17.4K]

The total bill for the Cunningham family was $29.26.

Step-by-step explanation:

Step 1; First we total up the cost of the food alone. Mr. Cunningham spent $7.50, Mrs. Cunningham spent $6.25, the 2 children spent $4.99 each which equals $9.98 for both. So the total of the food alone was $23.73.

Step 2; The sales tax rate was 7.25% of the $23.73. So we multiply 0.0725 (7.25%) with the $23.73 which equals $1.72. So the total bill was for $23.73 + $1.72 which equals $25.45.

Step 3; The Cunningham family left a 15% tip. This was 15% of the toal bill which was $25.45. So we multiply 0.15 (15%) with $25.45 which equals $3.81. So the total bill was food cost + service tax + tips which is $23.73 + $1.72 + $3.81 = $29.26

5 0
3 years ago
Assume a test for a disease has a probability 0.05 of incorrectly identifying an individual as infected (False Positive), and a
Nana76 [90]

Answer:

0.00002 = 0.002% probability of actually having the disease

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Positive test

Event B: Having the disease

Probability of having a positive test:

0.05 of 1 - 0.000001(false positive)

0.99 of 0.000001 positive. So

P(A) = 0.05*(1 - 0.000001) + 0.99*0.000001 = 0.05000094

Probability of a positive test and having the disease:

0.99 of 0.000001. So

P(A \cap B) = 0.99*0.000001 = 9.9 \times 10^{-7}

What is the probability of actually having the disease

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{9.9 \times 10^{-7}}{0.05000094} = 0.00002

0.00002 = 0.002% probability of actually having the disease

6 0
3 years ago
Find the surface area of the cone
babymother [125]
32.2 is the answer
hope it helps
8 0
3 years ago
In a certain town 215 boys and 265 girls were born last year. What is the probability that a child chosen at random from among t
Tanya [424]

Answer:

What I think it is from 5 through 25 or something like that

5 0
3 years ago
The average of your three quiz grades is 17 points. Two of your quiz grades are 14 and 19 points. Write and solve an equation to
Dennis_Churaev [7]
(14+19+x)/3=17
x= 18
the first is the equation and the second it the answer
5 0
4 years ago
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