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Vinvika [58]
3 years ago
10

The table shows how far a distance runner has traveled since the race began. What is her average speed, in miles per hour, durin

g the interval 0.75 to 1.00 hours?
Mathematics
2 answers:
grigory [225]3 years ago
8 0

Answer:

5\ \frac{miles}{hour}

Step-by-step explanation:

<u><em>The complete question is</em></u>

The table shows how far a distance runner has traveled since the race began. What is her average speed, in miles per hour, during the interval 0.75 to 1.00 hours?

Time Elapsed (Hours) Miles Traveled (Miles)

0.50 2.00

0.75 3.50

1.00 4.75

Let

x ----> the time in hours

f(x) ----> the distance traveled in miles

so

we have

x (Hours)   0.50   0.75  1.00

f(x) (Miles)  2.00   3.50  4.75

we know that

To find the average speed, we divide the change in the output value by the change in the input value

the average speed is equal to

\frac{f(b)-f(a)}{b-a}

In this problem we have

a=0.75

b=1.00

f(a)=f(0.75)=3.50  

f(b)=f(1.00)=4.75

Substitute

\frac{4.75-3.50}{1-0.75}

\frac{1.25}{0.25}

5\ \frac{miles}{hour}

Ierofanga [76]3 years ago
8 0

Answer:5.00 miles per hour

Step-by-step explanation:

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I need help with this
olga2289 [7]

Answer:

x = 24.52

Step-by-step explanation:

Since this is a right triangle, use Pythagorean Theorem to solve for the hypotenuse.

Pythagorean Theorem: a² + b² = c²

24² + 5² = x²

576 + 25 = x²

601 = x²

√601 = x

4 0
3 years ago
You toss three 6-sided dice and record the sum of the three faces facing up. a) Describe precisely a sample space S for this exp
timama [110]

Answer:

a.)Sample space means all possible outcomes, since we know the dice are 6 face, then the sample space becomes all possible outcomes when we toss the die.

b.) 9/216

c.) 9/216

d.) 212/216

Step-by-step explanation:

Sample space means all possible outcomes, since we know the dice are 6 face, then the sample space becomes all possible outcomes when we toss the die.

[1,1,1] [1,1,2] [1,1,3] [1,1,4] [1,1,5] [1,1,6]

[1,2,1] [1,2,2] [1,2,3] [1,2,4] [1,2,5] [1,2,6],

[1,3,1] [1,3,2] [1,3,3] [1,3,4] [1,3,5] [1,3,6]

[1,4,1] [1,4,2] [1,4,3] [1,4,4] [1,4,5] [1,4,6]

[1,5,1] [1,5,2] [1,5,3] [1,5,4] [1,5,5] [1,5,6]

[1,6,1] [1,6,2] [1,6,3] [1,6,4] [1,6,5] [1,6,6]

[2,1,1] [2,1,2] [2,1,3] [2,1,4] [2,1,5] [2,1,6]

[2,2,1] [2,2,2] [2,2,3] [2,2,4] [2,2,5] [2,2,6],

[2,3,1] [2,3,2] [2,3,3] [2,3,4] [2,3,5] [2,3,6]

[2,4,1] [2,4,2] [2,4,3] [2,4,4] [2,4,5] [2,4,6]

[2,5,1] [2,5,2] [2,5,3] [2,5,4] [2,5,5] [2,5,6]

[2,6,1] [2,6,2] [2,6,3] [2,6,4] [2,6,5] [2,6,6]

[3,1,1] [3,1,2] [3,1,3] [3,1,4] [3,1,5] [3,1,6]

[3,2,1] [3,2,2] [3,2,3] [3,2,4] [3,2,5] [3,2,6],

[3,3,1] [3,3,2] [3,3,3] [3,3,4] [3,3,5] [3,3,6]

[3,4,1] [3,4,2] [3,4,3] [3,4,4] [3,4,5] [3,4,6]

[3,5,1] [3,5,2] [3,5,3] [3,5,4] [3,5,5] [3,5,6]

[3,6,1] [3,6,2] [3,6,3] [3,6,4] [3,6,5] [3,6,6]

[4,1,1] [4,1,2] [4,1,3] [4,1,4] [4,1,5] [4,1,6]

[4,2,1] [4,2,2] [4,2,3] [4,2,4] [4,2,5] [4,2,6],

[4,3,1] [4,3,2] [4,3,3] [4,3,4] [4,3,5] [4,3,6]

[4,4,1] [4,4,2] [4,4,3] [4,4,4] [4,4,5] [4,4,6]

[4,5,1] [4,5,2] [4,5,3] [4,5,4] [4,5,5] [4,5,6]

[4,6,1] [4,6,2] [4,6,3] [4,6,4] [4,6,5] [4,6,6]

[5,1,1] [5,1,2] [5,1,3] [5,1,4] [5,1,5] [5,1,6]

[5,2,1] [5,2,2] [5,2,3] [5,2,4] [5,2,5] [5,2,6],

[5,3,1] [5,3,2] [5,3,3] [5,3,4] [5,3,5] [5,3,6]

[5,4,1] [5,4,2] [5,4,3] [5,4,4] [5,4,5] [5,4,6]

[5,5,1] [5,5,2] [5,5,3] [5,5,4] [5,5,5] [5,5,6]

[5,6,1] [5,6,2] [5,6,3] [5,6,4] [5,6,5] [5,6,6]

[6,1,1] [6,1,2] [6,1,3] [6,1,4] [6,1,5] [6,1,6]

[6,2,1] [6,2,2] [6,2,3] [6,2,4] [6,2,5] [6,2,6],

[6,3,1] [6,3,2] [6,3,3] [6,3,4] [6,3,5] [6,3,6]

[6,4,1] [6,4,2] [6,4,3] [6,4,4] [6,4,5] [6,4,6]

[6,5,1] [6,5,2] [6,5,3] [6,5,4] [6,5,5] [6,5,6]

[6,6,1] [6,6,2] [6,6,3] [6,6,4] [6,6,5] [6,6,6]

b.) Probability that the sum is 16 or more is

Pr[4,6,6] + pr[5,5,6] + pr [ 5,6,5] + pr [5,6,6] + pr [6,5,5] + pr [6,5,6] + pr [6,6,4] + pr[6,6,5] + pr [6,6,6]

Becomes:

[1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ = 9/216

Probability that the sum is 4 or 5 is

Pr [ 1,1,2] or pr[1,2,2] or pr [1,1,3] or pr [1,2,1] or pr[2,1,2] or pr[1,3,1] or pr[3,1,1] or pr[2,1,1] or pr[2,2,1]

Becomes:

[1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ = 9/216

Probability that the sum is less than 17

We take it as:

1- probability that the sum is 17 and above.

Now probability that the sum is 17 and above becomes

pr[5,6,6] or pr[6,5,6] or pr[6,6,5] or pr[6,6,6]

= [1/6]³ + [1/6]³ + [1/6]³ + [1/6]³ = 4/216

Hence, probability that the sum is less than 17 becomes:

1-4/216 = 212/216.

3 0
3 years ago
Given that BD is the median of AABC and that AABC is isosceles, congruence postulate SSS can be used to prove which of the follo
marusya05 [52]

Answer:

B.

Step-by-step explanation:

Because of isosceles triangle, AB = CB.

Because of the median, AD = CD.

Because of congruence of segments being reflexive, BD = BD.

By SSS, the triangles BAD and BCD are congruent.

Answer: B.

7 0
3 years ago
A scale drawing of an apartment is shown. What are the actual dimensions of the Living Space?
OLEGan [10]

Answer: 12 Centimeter

Step-by-step explanation:

3 by 4 multiplied by 2

3cm x 4cm x 2

= 12cm x 2

= 24cm

3 0
4 years ago
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What is the total surface area of the triangular prism shown?
Sergeu [11.5K]

Answer:

64

Step-by-step explanation:

6 0
3 years ago
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