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3 years ago

Hmmmmmmmmmmmmmmmmmmmmmmm i dont know

Mathematics

1 answer:

saw5 [17]3 years ago

4 0

Boopity boop boop boop boop booooop

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Im doing a test review and I have no idea how to do this

Arisa [49]

Answer:

94.3?

Step-by-step explanation:

Well, the straight line is 180 degrees, there is a 90 degree square, and a 13 degree angle. So add them up and divide by three?

6 0

4 years ago

Find the mean, median, mode, and range for the data set:<br> 8,6,4,4,5,6,4,9,4

xenn [34]

Answer:

Mean : 5.75

Median : 5.5

Mode : 4

Step-by-step explanation:

Hope this helps

4 0

3 years ago

Can Someone help with problems 2,3,4. Please

Ghella [55]

Number 2 is 990 because the numbers are 9, 10, and 11.

5 0

3 years ago

Todd plans to run at least 5 miles each week for his health. has a circular route in the neighborhood to run. Once around that r

adoni [48]

Answer: Yes

Step-by-step explanation:

Given

Once around park measures 420 yards

Todd needs to complete 5 miles

He runs 30 times during the week

We know, $1\ mile=1760\ yards$

for 30 rounds, distance is

$\Rightarrow 30\times 420=12,600\ \text{yards}$

In miles, it is

$\Rightarrow 12,600\equiv \dfrac{12,600}{1760}\\\Rightarrow 7.159\approx 7.16\ \text{miles}$

As he runs more than the target. Therefore, he completed his target of 5 miles.

3 0

3 years ago

4 men and 6 women are ranked according to their scores on an exam. Assume that no two scores are alike, and that all 10! 10 ! po

anygoal [31]

Answer:

a) $P(X=2)=\frac{2}{15}$

b) $P(X=3)=\frac{1}{30}$

c) P(X=6)=0

d) P(X=9)=0

Step-by-step explanation:

We know that are 4 men and 6 women are ranked according to their scores on an exam. X = 1 indicates that a man achieved the highest score on the exam.

a) We calculate P(X=2).

We calculate the number of possible combinations

$C^{10}_{2}=\frac{10!}{2! (10-2)!}=\frac{10\cdot 9\cdot 8!}{2\cdot 1 \cdot 8!}=45$

We calculate the number of favorable combinations

$C_2^4=\frac{4!}{2!(4-2)!}=6$

We get that is

$\boxed{P(X=2)=\frac{6}{45}=\frac{2}{15}}$

b) We calculate P(X=3).

We calculate the number of possible combinations

$C^{10}_{3}=\frac{10!}{3! (10-3)!}=\frac{10\cdot 9\cdot 8\cdot 7!}{3\cdot2\cdot 1 \cdot 7!}=120$

We calculate the number of favorable combinations

$C_3^4=\frac{4!}{3!(4-3)!}=4$

We get that is

$\boxed{P(X=3)=\frac{4}{120}=\frac{1}{30}}$

c) We calculate P(X=6). This case is not possible because 6 men cannot be selected because we have been given 4 men.

We conclude P(X=6)=0.

d) We calculate P(X=9). This case is not possible because 9 men cannot be selected because we have been given 4 men.

We conclude P(X=9)=0.

5 0

3 years ago