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3 years ago

What is the result of -1/8 times 16 2/3?

Mathematics

1 answer:

Diano4ka-milaya [45]3 years ago

5 0

I hope this helps you

16 2/3= 16.3+2/3=50/3

-1/8.50/3

-25/12

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A tune up automotive facility marks up its part by 25% suppose that the facility charges its customers $1.75 for each spark plug

ss7ja [257]

Answer:

$1.4

Step-by-step explanation:

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3 years ago

If f(x) = x2 – 2x and g(x) = 6x + 4, for which value of x does (f + g)(x) = 0?

Hatshy [7]

Answer:

x = - 2

Step-by-step explanation:

(f + g)(x)

= f(x) + g(x)

= x² - 2x + 6x + 4

= x² + 4x + 4

Equating to zero

x² + 4x + 4 = 0

(x + 2)² = 0

x + 2 = 0

x = - 2

6 0

3 years ago

As Saturn revolves around the sun, it travels at a speed of approximately 6 miles per second. Convert this speed to miles per mi

leva [86]

Answer:

The answer is 21,600. If it travels 6 miles per second that in a minute it would travel 360 miles. You multilpy 360 times 6 and you get your answer of 21,600 miles.

Step-by-step explanation:

5 0

3 years ago

BaLLatris [955]

<h3>Answer:</h3>

x=30

<h3>Solution:</h3>

First, use the distributive property:
8x-3(2x-4)=3(x-6)
8x-6x+12=3x-18
Simplify:
2x+12=3x-18
2x-3x=-12-18
-x=-30
x=30

Hope it helps.

Do comment if you have any query.

8 0

2 years ago

Suppose a particular type of cancer has a 0.9% incidence rate. Let D be the event that a person has this type of cancer, therefo

natita [175]

Answer:

There is a 12.13% probability that the person actually does have cancer.

Step-by-step explanation:

We have these following probabilities.

A 0.9% probability of a person having cancer

A 99.1% probability of a person not having cancer.

If a person has cancer, she has a 91% probability of being diagnosticated.

If a person does not have cancer, she has a 6% probability of being diagnosticated.

The question can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem we have the following question

What is the probability that the person has cancer, given that she was diagnosticated?

P(B) is the probability of the person having cancer, so $P(B) = 0.009$

P(A/B) is the probability that the person being diagnosticated, given that she has cancer. So $P(A/B) = 0.91$

P(A) is the probability of the person being diagnosticated. If she has cancer, there is a 91% probability that she was diagnosticard. There is also a 6% probability of a person without cancer being diagnosticated. So

$P(A) = 0.009*0.91 + 0.06*0.991 = 0.06765$

What is the probability that the person actually does have cancer?

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.91*0.009}{0.0675} = 0.1213$

There is a 12.13% probability that the person actually does have cancer.

3 0

3 years ago