MATHEMATICS COLLEGE

Find the missing Length X
Help me !!!!
A.12
B.11
C.5

Answers

Answer 1

Answer:

Answer:

Step-by-step explanation:

Δ PQR ~ ΔJKL ⇒ PQ : JK = PR : JL

9 : 6 = x : 8

6x = 72 ⇒ x = 12

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Using division to determine which of the following numbers are factors of 351
2
3
5
7
35

so which are factors and which are not

Answers

Answer:

Factors:

1, 5, 7, 35

Non-Factors:

2, 3

Step-by-step explanation:

A study was conducted to investigate the relationship between soda consumptions (servings per day) and weight gain (kgs) over a 2 year period and yielded the following linear regression results: weight gain over a 2 year period (kgs) = 4.5 + 0.25(servings of soda per day) what is the predicted weight gain over a 2 year period for someone that drinks 4 servings of soda per day?

Answers

Answer:

5.5kg

Step-by-step explanation:

Given that in a study conducted to investigate the relationship between soda consumptions (servings per day) and weight gain (kgs) over a 2 year period and yielded the following linear regression results:

Let y = weight gains over a two year period (kg) and

x = servings of soda per day

Then the regression equation estimated is

$4.5+0.25x =y$

Using this we can find unknown y for a given x.

When x =4 we get y by substituting x =4 in the equation

$y (4) = 1+4.5 = 5.5$

Weight gain = 5.5 kg.

At a point on the ground 20 feet from a building, a surveyor observes the angle of inclination to the top of the building to be pi/3 radians. How tall is the building?

Answers

Answer:

34.64 ft

Step-by-step explanation:

Distance from the building = 20 ft

Angle of inclination = π/3 radians

The tangent of the angle of inclination must equal the height of the building divided by the distance of the observer from the building:

$tan( \pi/3) = (h)/(20) \nh = 20*1.73205\nh=34.64\ ft$

The building is 34.64 ft tall

Which functions have a maximum value greater than the maximum of the function g(x) = –(x + 3)2 – 4? Check all that apply. f(x) = –(x + 1)2 – 2 f(x) = –|x + 4| – 5 f(x) = –|2x| + 3

Answers

Answer:

Step-by-step explanation:

Please help!!What is the value of x? Enter your answer in the box. x = NOTE: Image not drawn to scale. Triangle G E H with segment E D such that D is on segment G H, between G and H. Angle G E D is congruent to angle D E H. E G equals 44.8 millimeters, G D equals left parenthesis x plus 4 right parenthesis millimeters, D H equals 35 millimeters, and E H equals 56 millimeters.

Answers

Answer:

The value of x is 24.

Step-by-step explanation:

Given information: In ΔGHE, ED is angle bisector, EG=44.8 millimeters, GD=(x+4) millimeters, DH=35 millimeters, and EH=56 millimeters.

According to the angle bisector theorem, an angle bisector divide the opposite side into two segments that are proportional to the other two sides of the triangle.

In ΔGHE, ED is angle bisector, By using angle bisector theorem, we get

$(GD)/(DH)=(EG)/(EH)$

$(x+4)/(35)=(44.8)/(56)$

Multiply both the sides by 35.

$x+4=(44.8)/(56)* 35$

$x+4=28$

Subtract 4 from both the sides.

$x=28-4$

$x=24$

Therefore the value of x is 24.

Answer:

x = 24

Step-by-step explanation:

The segments on either side of an angle bisector are proportional:

(x +4)/44.8 = 35/56

x +4 = 44.8·(35/56) = 28 . . . . multiply by 44.8

x = 24 . . . . . subtract 4

It is important that face masks used by firefighters be able to withstand high temperatures because firefighters commonly work in temperatures of 200-500 degrees. In a test of one type of mask, 24 of 55 were found to have their lenses pop out at 325 degrees. Construct and interpret a 93% confidence interval for the true proportion of masks of this type whose lenses would pop out at 325 degrees.

Answers

Answer:

The 93% confidence interval for the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574). This means that we are 93% sure that the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.