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astraxan [27]
3 years ago
9

Over the weekend, the All Smiles Run had 16,478 participants for the Half Marathon. The All Smiles 10K had 6,924 participants, a

nd the 5-Mile Walk had 3,067 participants.
How many more people participated in the Half Marathon than in the 10K race and the 5-Mile Walk combined?



A.


6,487


B.


6,587


C.


9,991


D.


12,621
Mathematics
1 answer:
Kitty [74]3 years ago
6 0
 combine the number of participants in the 10k and 5-mile
6924+3067=9991
Now subtract the half marathon from 9991
16,478-9991=6487
Your answer would be A.
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A right triangle has one leg with a length of 48 and a hypotenuse with a length of 50. HELPPPPPP
makkiz [27]

Answer:

14 is the length of the missing side.

Step-by-step explanation:

Use pythagorean theorem.

48^{2}+b^{2}=50^{2}

2304+b^{2}=2500

\sqrt{b^{2}}=\sqrt{196}

b=14

3 0
3 years ago
Read 2 more answers
Help with this problem
natita [175]
Y = 6x - 4

basically, we just sub answer choices in to see which ones make the equation true.

answer is : (0,-4) , (1,2) , (3,14)
3 0
3 years ago
Mr. Whitman gave a test to his 32 math students. He felt the results were low, so he retested them one week later. The results o
Liono4ka [1.6K]

The number of student more passed the retake, when the students need a 65 in order to pass the test is 8.

<h3>What is box and whiskers plot?</h3>

Box plot is the way of representation of data which gives the graphical image of the data set to understand better. In this the data is represented with the help of quartile.

  • A.The difference in the median grade-

The median of first test is 60 while the second test is 80. Thus, the difference is,

d=80-60\\d=20

  • B. Compare the percent of students who got at least a 60 on the original test to those that got at least a 60 on the retake.

The student, who got at least 60 in first test, falls in the second half of the plot. Thus, the 50% of students got atleast 60% marks. In the retake test, all 100% student got at least 60 on test.

  • C. If the students need a 65 in order to pass the test, how many more passed the retake?

In the first test 1/4 are passed while in second test 3/4 students are passed and got the marks 65 or more than this, Thus the number of student pass more in this test,

n=(\dfrac{3}{4}-\dfrac{1}{2})\times 32\\n=8

Thus, the number of student more passed the retake, when the students need a 65 in order to pass the test is 8.

Learn more about the box and whiskers plot here;

brainly.com/question/9559392

#SPJ1

8 0
2 years ago
A, B, and C are the vertices of one triangle. D, E, and F are the vertices of another triangle. m∠A = 50, m∠B = 10, m∠E = 40, m∠
adell [148]
The fourth one seems like its the right one 
3 0
3 years ago
A supplier delivers an order for 20 electric toothbrushes to a store. By accident, three of the electric toothbrushes are defect
Vladimir79 [104]

The probability that the first two electric toothbrushes sold are defective is 0.016.

The probability of an event, say E occurring is:

P(E)=\frac{n(E)}{N}

Here,

n (E) = favorable outcomes

N = total number of outcomes

Let X = the number of defective electric toothbrushes sold.

The number of electric toothbrushes that were delivered to a store is n = 20.

The number of defective electric toothbrushes is x = 3.

The number of ways to select two toothbrushes to sell from the 20 toothbrushes is:

(\left {{20} \atop {2}} \right. )=\frac{20!}{2!(20-2)!} =\frac{20!}{2!18!} =\frac{20*19*18!}{2!*18!} = 190

The number of ways to select two defective toothbrushes to sell from the 3 defective toothbrushes is:

(\left {{3} \atop {2}} \right. )=\frac{3!}{2!(3-2)!} =\frac{3!}{2!1!} =\frac{3*2!}{1!2!} =3

Compute the probability that the first two electric toothbrushes sold are defective as follows:

P (Selling 2 defective toothbrushes) = Favorable outcomes ÷ Total no. of outcomes

\frac{3}{190}\\ =0.01579\\=0.016

Thus, the probability that the first two electric toothbrushes sold are defective is 0.016.

Learn more about probability here brainly.com/question/27474070

#SPJ4

4 0
2 years ago
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