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2 years ago

Question explains itself in the picture.

Mathematics

1 answer:

borishaifa [10]2 years ago

4 0

Answer:

30 cm

Step-by-step explanation:

Area of a triangle formula:

A = 0.5bh

Given:

b = 12

c = 5

Work:

A = 0.5bh

A = 0.5(12)(5)

A = 6(5)

A = 30

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What data is the table referring to?

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3 years ago

A direct variation function contains the points (2, 14) and (4, 28). Which equation represents the function?

vitfil [10]

A direct variation has a constant slope, i.e. (y/x)= constant.

Both of the given two points give a slope of (y/x)=14/2=28/4=7, so the equation of the function is
y/x=7, or simply
y=7x

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2 years ago

The graph of g(x)=(0.75)x+2 is shown.

Andrei [34K]

we are given

$f(x)=(0.75)^x+2$

For finding asymptote , we can find limit

$\lim_{x \to \infty} f(x)= \lim_{x \to \infty}((0.75)^x+2)$

$\lim_{x \to \infty} f(x)= \lim_{x \to \infty} (0.75)^x+\lim_{x \to \infty} 2)$

now, we can solve it

$\lim_{x \to \infty} f(x)= 0+2$

$\lim_{x \to \infty} f(x)= 2$

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$y= 2$ .............Answer

6 0

2 years ago

A picture frame can hold a Photograph that is 16 inches long and 12 inches wide. What is the area of the photo that fits the fra

TEA [102]

Answer:

16x12=192

Step-by-step explanation:

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8 0

2 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