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ICE Princess25 [194]
2 years ago
11

NEED HELP ASAP PLEASE!!!!

Mathematics
1 answer:
victus00 [196]2 years ago
7 0
Try 11 because 36 plus 43 plus 90(right angle) is 169. The total degree of yeh triangle is 180 so 180-169 is 11
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Jonas jogged up the hill at an average rate of of a mile per minute and then walked down the hill at an average rate of of a mil
jenyasd209 [6]
The missing value depends on the speed of jonas to total miles  covered by jonas.......if we were to assume that he jogged up the hill at a distance of 15minutes  then the trip took him 42 min in total so the remaining travel time would be 27 minutes .....lets say he is walking down the hill at .5 miles per min,.. you shpuld multiply by 27 by .5 which is 13.5
4 0
3 years ago
Kaylee drove 75.5 miles in 54 minutes,if she continues to drive at the same rate, how long will it take her to drive 250 miles?
Crazy boy [7]

Answer:

179 minutes

Step-by-step explanation:

  1. Divide both sides by 75.5 to see how many minutes it would take to drive 1 mile. (54 / 75.5) =108/151
  2. She can drive 1 mile in 108/151 minutes
  3. Times by 250. (108/151 * 250) 178.8
  4. 179 to nearest minute
7 0
3 years ago
The area of the shaded segment is 100cm^2. Calculate the value of r.
Reil [10]
Hello, 

The formula for finding the area of a circular region is: A=  \frac{ \alpha *r^{2} }{2}

then:
A_{1} = \frac{80*r^{2} }{2}

With the two radius it is formed an isosceles triangle, so, we must obtain its area, but first we obtain the height and the base.

cos(40)= \frac{h}{r}  \\  \\ h= r*cos(40)\\ \\ \\ sen(40)= \frac{b}{r} \\ \\ b=r*sen(40)

Now we can find its area:
A_{2}=2* \frac{b*h}{2}  \\  \\ A_{2}= [r*sen(40)][r*cos(40)]\\  \\A_{2}= r^{2}*sen(40)*cos(40)

The subtraction of the two areas is 100cm^2, then:

A_{1}-A_{2}=100cm^{2} \\ (40*r^{2})-(r^{2}*sen(40)*cos(40) )=100cm^{2} \\ 39.51r^{2}=100cm^{2} \\ r^{2}=2.53cm^{2} \\ r=1.59cm

Answer: r= 1.59cm
7 0
2 years ago
Read 2 more answers
According to the data, the mean quantitative score on a standardized test for female college-bound high school seniors was 500.
Sveta_85 [38]

Answer:

6.68% of the female college-bound high school seniors had scores above 575.  

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 500

Standard Deviation, σ = 50

We are given that the distribution of score is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

P(scores above 575)

P(x > 575)

P( x > 575) = P( z > \displaystyle\frac{575 - 500}{50}) = P(z > 1.5)

= 1 - P(z \leq 1.5)

Calculation the value from standard normal z table, we have,  

P(x > 575) = 1 - 0.9332= 0.0668 = 6.68\%

6.68% of the female college-bound high school seniors had scores above 575.

7 0
3 years ago
Evaluate the expression using the given value 3n+7,n=12
balandron [24]
3x12=36
36+7= 43

The answer is 43
6 0
1 year ago
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