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3 years ago

Plz help I need help plz

Mathematics

1 answer:

dem82 [27]3 years ago

4 0

Answer:

33/4 or 8 1/4

Step-by-step explanation:

-3 2/3 • (-2 1/4)

Step 1. Convert each into improper fractions
(Reply for details)

-11/3 • (-9/4)

Step 2. Multiply
(Reply for details)

99/12

Step 3. Simplify
(Reply for details)

33/4 or 8 1/4

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A club raised 175% of its goal for a charity. The club raises $875. What was the goal?

charle [14.2K]

Answer: The goal was $500

======================================================

Work Shown:

x% = x/100

175% = 175/100 = 1.75

Let g be the goal, which is the amount of money the club wanted to raise

175% of goal = 175% of g = 1.75g

The expression 1.75g represents how much money was actually raised, which was $875. Set the two expressions equal to each other. Solve for g

1.75g = 875

g = 875/1.75 ....... divide both sides by 1.75

g = 500

The club's goal was to raise $500

Note how 75% of $500 is 0.75*500 = 375

When they raised 175% of the goal, this means they went 75% overboard and added on 375 additional dollars (on top of the 500 they wanted). So they got to 500+375 = 875 which lines up with the instructions. This helps verify the answer.

Or we can see that 1.75*g = 1.75*500 = 875 which helps confirm the answer as well.

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3 years ago

kyle took at total of 8 quizzes in the first 4 weeks of school. How many weeks of school will he have to attend before he will h

Elden [556K]

Answer:

6 weeks

Step-by-step explanation:

8 quizzes = 4 weeks

So 2 quizzes/week

It will take her 12/2 weeks (6) to finish 12 quizzes

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2 years ago

Solve for x in simplest form. 2=1/2(7x-4)

Sladkaya [172]

Answer:1.1428 is the answer

Step-by-step explanation: Distribute using (Pemdas) 2nd find x alone and divide by 4

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3 years ago

Does arkansas lie south of 40 degrees latitude

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All of Arkansas is between 33 degrees and 36.5 degrees lol

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3 years ago

A statistician is testing the null hypothesis that exactly half of all engineers will still be in the profession 10 years after

lana [24]

Answer:

95% confidence interval estimate for the proportion of engineers remaining in the profession is [0.486 , 0.624].

(a) Lower Limit = 0.486

(b) Upper Limit = 0.624

Step-by-step explanation:

We are given that a statistician is testing the null hypothesis that exactly half of all engineers will still be in the profession 10 years after receiving their bachelor's.

She took a random sample of 200 graduates from the class of 1979 and determined their occupations in 1989. She found that 111 persons were still employed primarily as engineers.

Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of persons who were still employed primarily as engineers = $\frac{111}{200}$ = 0.555

n = sample of graduates = 200

p = population proportion of engineers

Here for constructing 95% confidence interval we have used One-sample z proportion test statistics.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of

significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.555-1.96 \times {\sqrt{\frac{0.555(1-0.555)}{200} } }$ , $0.555+1.96 \times {\sqrt{\frac{0.555(1-0.555)}{200} } }$ ]

= [0.486 , 0.624]

Therefore, 95% confidence interval for the estimate for the proportion of engineers remaining in the profession is [0.486 , 0.624].

7 0

3 years ago