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3 years ago

What is the mode of the data?

Mathematics

1 answer:

Lesechka [4]3 years ago

7 0

Answer:

0, 5, and 9

Step-by-step explanation:

Mode is (are) the number(s) that is (are) repeated the most times. Lets rewrite the numbers in clusters to see which number(s) are repeated the most times:

0 0 0

1 1

2 2

3 3

5 5 5

9 9 9

The numbers 0, 5, and 9 are repeated 3 times each. So these three numbers are the mode.

Hope it helps!

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Tiffany is filling her daughter's pool with water from a hose she can fill the water at a rate of 1/10 gallons/second. Create at

s2008m [1.1K]

Answer:

Three equivalent ratios that are associated with this rate would be 6 gallons/1min, 360gallons/1 hour and 720 gallons/2hours.

I'm sorry but I can't explain or show how to use a double number line for this question.

Step-by-step explanation:

4 0

2 years ago

The volume of a box is 400in3. if the height is 10in and the length is 8in, what is the width of the box (volume=l•w•h)

tankabanditka [31]

10•8=80
so something times 80 will equal 400
80•5=400

6 0

3 years ago

What is the cosecant of angle C?

pishuonlain [190]

Cosecant = hypotenuse / opposite side ( that is 1 / sine)
csc C = AC / 8

AC = sqrt (6^2 + 8^2) = sqrt 100 = 10

so cosecant of C is 10/8 = 1.25

5 0

3 years ago

Plzzzzz help me with this

ehidna [41]

Answer:

We conclude that we must add 4² or 16 to complete the square.

Hence, option C i.e. 16 is the correct answer.

Step-by-step explanation:

We know that the perfect square formula is

(a + b)² = a² + b² + 2ab

Given the equation

$x^2+8x$

Let us add 4² or 16 in the equation

$x^2+8x+4^2$

$x^2+\:2\left(1\right)\left(4\right)x+4^2$

as (a + b)² = a² + b² + 2ab, so

$x^2+8x+4^2=\left(x+4\right)^2$

Therefore, we conclude that we must add 4² or 16 to complete the square.

Hence, option C i.e. 16 is the correct answer.

4 0

3 years ago

Read 2 more answers

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

5 0

2 years ago