3a. The graph shows a school's profit P for selling x lunches on one day. PART A:

Nadusha1986 [10]

Answer:

2.6

Step-by-step explanation:

i hope this helps bro because i know this is the right answer because i think so and im pretty sure im right because I think so

5 0

2 years ago

Read 2 more answers

What is the correct factorization of:<br> x? - 11x + 18

Dominik [7]

9514 1404 393

Answer:

x² -11x +18 = (x -9)(x -2)

Step-by-step explanation:

The x-coefficient is the sum of the constants in the binomial factors; the constant term is their product. It usually works well to look for factors of the constant that have the right sum. Here, both must be negative.

18 = (-1)(-18) = (-2)(-9) = (-3)(-6)

The factors -2 and -9 have a sum of -11, so those are the binomial constants of interest. The factorization is ...

x² -11x +18 = (x -9)(x -2)

8 0

2 years ago

Mason borrowed 1,200 from a friend to buy a new hot tub, His friend does't charge any interest and mason makes 50 payments eavh

balu736 [363]

You can identify the slope and y-intersept. The slope is 50 and the y-intercept is (0, 1200). You could also do the x-intercept, which would be the number of months passed when he pays off the debt fully. The x intercept would be (24,0)

5 0

2 years ago

15. Sensei's Judo Club has a competition for

marta [7]

Answer:

6 grapples

Step-by-step explanation:

x = number of wins

y = number of ties

y = 15 - x

48 = 5x + 2y

48 = 5x + 2(15 - x)

48 = 5x + 30 - 2x

48 = 3x + 30

18 = 3x

x = 6

3 0

2 years ago

A survey report states that 70% of adult women visit their doctors for a physical examination at least once in two years. If 20

irakobra [83]

Answer:

a) 0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

b) 0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

Step-by-step explanation:

For each women, there are only two possible outcomes. Either they visit their doctors for a physical examination at least once in two years, or they do not. The probability of a woman visiting their doctor at least once in this period is independent of any other women. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

70% of adult women visit their doctors for a physical examination at least once in two years.

This means that $p = 0.7$

20 adult women

This means that $n = 20$

(a) Fewer than 14 of them have had a physical examination in the past two years.

This is:

$P(X < 14) = 1 - P(X \geq 14)$

In which

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

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

$P(X = 14) = C_{20,14}.(0.7)^{14}.(0.3)^{6} = 0.1916$

$P(X = 15) = C_{20,15}.(0.7)^{15}.(0.3)^{5} = 0.1789$

$P(X = 16) = C_{20,16}.(0.7)^{16}.(0.3)^{4} = 0.1304$

$P(X = 17) = C_{20,14}.(0.7)^{17}.(0.3)^{3} = 0.0716$

$P(X = 18) = C_{20,18}.(0.7)^{18}.(0.3)^{2} = 0.0278$

$P(X = 19) = C_{20,19}.(0.7)^{19}.(0.3)^{1} = 0.0068$

$P(X = 20) = C_{20,20}.(0.7)^{20}.(0.3)^{0} = 0.0008$

So

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1916 + 0.1789 + 0.1304 + 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.6079$

$P(X < 14) = 1 - P(X \geq 14) = 1 - 0.6079 = 0.3921$

0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

(b) At least 17 of them have had a physical examination in the past two years

$P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

From the values found in item (a).

$P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.107$

0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

6 0

1 year ago