All categories

3 years ago

Passage : ( -5/6) + 2/4

Mathematics

1 answer:

Ulleksa [173]3 years ago

4 0

To add fractions, find the LCD and then combine.
Exact Form:

- 1/3 ( write as fraction)

Decimal Form:

-0.3

You might be interested in

What is the volume of the rectangular prism? A prism has a length of 8 inches, width of 2 inches, and height of 12 and one-half

Arte-miy333 [17]

Answer:4

Step-by-step explanation:because 4

4 0

3 years ago

Read 2 more answers

During the Playoff the schools football team has on average 30% their games than during the regular season. The average for this

exis [7]

Answer:12000

Step-by-step explanation:

During the Playoff the schools football team has on average 30% their games than during the regular season

The average for this year's playoffs attendance was 15,600 fans

To get the number of fans during a regular season goes thus:

Since there is an increase of 30%, we add 100% to 30% which gives:

100% + 30% = 130%

= 15600 ÷ 130%

= 15600 ÷ 130/100

= 15600 × 100/130

= 15600 × 0.7692

= 12000

There are 12000 fans during a regular season

5 0

3 years ago

Read 2 more answers

A new test to detect TB has been designed. It is estimated that 88% of people taking this test have the disease. The test detect

Elodia [21]

Answer:

Correct option: (a) 0.1452

Step-by-step explanation:

The new test designed for detecting TB is being analysed.

Denote the events as follows:

<em>D</em> = a person has the disease

<em>X</em> = the test is positive.

The information provided is:

$P(D)=0.88\\P(X|D)=0.97\\P(X^{c}|D^{c})=0.99$

Compute the probability that a person does not have the disease as follows:

$P(D^{c})=1-P(D)=1-0.88=0.12$

The probability of a person not having the disease is 0.12.

Compute the probability that a randomly selected person is tested negative but does have the disease as follows:

$P(X^{c}\cap D)=P(X^{c}|D)P(D)\\=[1-P(X|D)]\times P(D)\\=[1-0.97]\times 0.88\\=0.03\times 0.88\\=0.0264$

Compute the probability that a randomly selected person is tested negative but does not have the disease as follows:

$P(X^{c}\cap D^{c})=P(X^{c}|D^{c})P(D^{c})\\=[1-P(X|D)]\times{1- P(D)]\\=0.99\times 0.12\\=0.1188$

Compute the probability that a randomly selected person is tested negative as follows:

$P(X^{c})=P(X^{c}\cap D)+P(X^{c}\cap D^{c})$

$=0.0264+0.1188\\=0.1452$

Thus, the probability of the test indicating that the person does not have the disease is 0.1452.

4 0

3 years ago

Help pls will mark brainliest

Viktor [21]

Answer:

kitchi kitch ya ya da da

Step-by-step explanation:

4 0

2 years ago

Lydia earned $9.75 per hour last summer. She worked 25 hours a week. About how much did she earn in 10 weeks.

Goryan [66]

Well $9.75 times 25 hours is $243.75.

Then you take that and multiply it by 10.

So she earned $2,437.5

5 0

3 years ago