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2 years ago

Segment AB is congruent to segment AB.This statement shows the blankkkkk property

Mathematics

1 answer:

kompoz [17]2 years ago

3 0

Reflexive Property, the property where the quantity equals itself.

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Are 30/80 and 24/64 proportional

Flura [38]

Answer:

Yes, they are proportional

Step-by-step explanation:

First, simplify each fraction

30/80 can be simplified by dividing each number by 10

30/80

= 3/8

24/64 can be simplified by dividing each number by 8

24/64

= 3/8

So, each fraction will simplify to 3/8, meaning that they are proportional.

5 0

2 years ago

Help!!!! f(x)=10x+3 g(x)= -7x-4

Gelneren [198K]

Answer:

8x + 7

Step-by-step explanation:

8 0

2 years ago

Explain how to do (2^-2/3^3)^4 i need to know how to do this.

cricket20 [7]

Answer:

((2^(-2))÷(3^3))^4=

Step-by-step explanation: expand

5 0

2 years ago

A website has a 826,140 hits last month what is the value of the 8 in 826,140

guapka [62]

The value of 8 is in the hundred thousands

hope this helps

5 0

3 years ago

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