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2 years ago

I WILL GIVE 20 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS AND EXPLAIN WHY THAT IS THE ANSWER

Mathematics

1 answer:

e-lub [12.9K]2 years ago

8 0

Answer:

x=11 and t=6

Step-by-step explanation:

If ABC and KLM are approximately equal, 44 degrees=4x and 18 inches=3t.

44/4=11

18/3=6

You might be interested in

Is 5 equal to -5 in math

elena-14-01-66 [18.8K]

Answer: No

Step-by-step explanation: If it is talking about the reciprocal then yes. Reciprocal is the opposite of your number.

Example: -4 IS 4 reciprocal.

6 0

3 years ago

Read 2 more answers

The lovin lemonade company sells a 4 gallon jug of lemonade for $24 the sweet and sour company sells an eight pack of 1 quart bo

lbvjy [14]

One way to compare is to find the price per gallon.
$24 for 4 gallons = 24÷4 = $6 per gallon

8 pack of 1 qt bottle give you 2 gallons (4qts= 1 gallon) So you get 2 gallons for $16
$16 ÷ 2 = $8.00 per gallon. Sweet and Sour has the higher unit price.
Even if we compare the price per quart we will get the same answer as to who is selling it at the higher unit price.
4gal for $24 = 16 quarts for $24 ( 1 gallon = 4 qts)
$24 ÷ 16 = $1.50 each (1 $\frac{1}2}$ )
vs.
8 quarts for $16 $16 ÷ 8 = $2.00 per qt. Sweet and sour still has the higher unit price

7 0

3 years ago

Read 2 more answers

Point J (-4, -6) and point K (4, 4) are located on a coordinate grid. Which measurement is closest to the distance between point

shepuryov [24]

Answer:

The points are sqrt(164) or 2sqrt(41) units apart.

Explanation:

You didn't provide the options in your question. But to find the distance, you need to use the distance formula.

d=sqrt((y2-y1)^2 + (x2-x1)^2)

d is the distance, and (x1, y1) and (x2, y2) are the points.

Use point J (-4,-6) for (x1,y1) and point K (4,4) for (x2, y2).

d=sqrt((4-(-6))^2 + (4-(-4))^2)

=sqrt(10^2 + 8^2)

=sqrt(100+64)

=sqrt(164)=2sqrt(41)

You can compare this to the answer options and find the closest. Hope I could help! :)

3 0

3 years ago

A. When x = -1, what is the value of y?

pogonyaev

Answer:

i think we need more context to this answer. I’d love to help no need to retype. You can just message me for help :)

Step-by-step explanation:

7 0

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

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

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

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

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

= [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