Ask question

Ask question

All categories

2 years ago

5

Below are two inequalities and the graphs of their lines without the shading. By imagining where the shading should be, identify

which point would satisfy BOTH inequalities

1 answer:

Sveta_85 [38]2 years ago

7 0

Answer:

(7,-8) it is i belive

Step-by-step explanation:

You might be interested in

What is domain and range in math

aksik [14]

Answer:

range is the difference between big and small numbers

7 0

2 years ago

Tell whether the triangle is a right triangle. Show Work.

Yakvenalex [24]

Answer:

This is not a right triangle because the hypotenuse is less than the two sides and aren't equal when following the Pythagorean thm

Step-by-step explanation:

To confirm if a triangle is a right triangle, we must use the Pythagorean thm

a^2 + b^2 = c^2

23^2 + 11.4^2 = 21.2^2

529 + 129.96 = 449.44

658.96 ≠ 449.44

7 0

3 years ago

Find the common difference of the arithmetic<br> sequence 4, 10, 16,

DIA [1.3K]

Answer:

10it is 10

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

(3x squared)to the power of -2

Nady [450]

$(3x^2)^{-2}\to\boxed{1/(3x^{2})^2\to\frac{1}{9x^4}}$

3 0

2 years ago

Read 2 more answers

A study showed that 25% of the students drive themselves to school. Based on the suggested probability, in a class of 18 student

Gwar [14]

Answer:

B) 0.283

Step-by-step explanation:

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.

25% of the students drive themselves to school.

This means that $p = 0.25$

Class of 18 students

This means that $n = 18$

What would be the probability that at least 6 students drive themselves to school?

This is

$P(X \geq 6) = 1 - P(X < 6)$

In which

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

So

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

$P(X = 0) = C_{15,0}.(0.25)^{0}.(0.75)^{18} = 0.006$

$P(X = 1) = C_{15,1}.(0.25)^{1}.(0.75)^{17} = 0.034$

$P(X = 2) = C_{15,2}.(0.25)^{2}.(0.75)^{16} = 0.096$

$P(X = 3) = C_{15,3}.(0.25)^{3}.(0.75)^{15} = 0.17$

$P(X = 4) = C_{15,4}.(0.25)^{4}.(0.75)^{14} = 0.213$

$P(X = 5) = C_{15,5}.(0.25)^{5}.(0.75)^{13} = 0.199$

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.006 + 0.034 + 0.096 + 0.17 + 0.213 + 0.199 = 0.718$

$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.718 = 0.282$

Closest option is B, just a small rounding difference.

4 0

2 years ago