All categories

3 years ago

HELP FAST pls pls pls pls

Mathematics

1 answer:

antiseptic1488 [7]3 years ago

3 0

Answer:

Stretch and shrink the graph of a function. To stretch or shrink the graph in the y direction, multiply or divide the output by a constant. 2f (x) is stretched in the y direction by a factor of 2, and f (x) is shrunk in the y direction by a factor of 2 (or stretched by a factor of ).

Step-by-step explanation:

hope this helps!

You might be interested in

One number is six more than a second number. four times the first is 10 less than 6 times the second. Find the numbers.

zavuch27 [327]

$\sf{\bold{\blue{\underline{\underline{Given}}}}}$

One number is six more than a second number.
four times the first is 10 less than 6 times the second.⠀⠀⠀⠀

$\sf{\bold{\red{\underline{\underline{To\:Find}}}}}$

Find the two numbers.⠀

$\sf{\bold{\purple{\underline{\underline{Solution}}}}}$

⠀⠀⠀⠀

In questions given,One number is six more than a second number.

if second number is a

first number is a+6.

and,

four times the first is 10 less than 6 times the second.

So,

4×(a+6)=6a-10

we have to solve this.

$\bold{4\times (a+6)=6a-10 }$

$\bold{4a+24=6a-10 }$

$\bold{6a-4a=24+10 }$

$\bold{2a=34 }$

$\bold{ a=\dfrac{34}{2} }$

$\bold{ a=17 }$

we,get our second number =a=17.

So,

first number =a+6=17+6=23

5 0

3 years ago

Please help me out with this question!!!

V125BC [204]

Answer:

100

Step-by-step explanation:

5 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:

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