Find the distance between the points (9, 8) and (5, 3).

Find P sophomore girl hint: p(a & b)/p(b)

Viktor [21]

Answer:

7/18

Step-by-step explanation:

P(sophomore girl) is 7 out of a total 30.

P(girl) is 18 out of 30.

(7/30) / (18/30) = 7/18

If you just look at the girls column you can see this immediately. A given is that we're looking for a girl (18 girls), then what is the chance that it's a sophomore. That is 7 out of 18.

7 0

2 years ago

Is the term (X + 2) a factor of the polynomial shown below?

QveST [7]

The correct answer is: No, the remainder is -114

Further explanation:

In order to check whether the given term is a factor of given polynomial or not we will put

x+2=0

x=-2

Given

$f(x)=x^3 - 10x^2 + 27^x - 12\\f(-2)=(-2)^3-10(-2)^2+27(-2)- 12\\=-8-10(4)-54-12\\=-8-40-54-12\\=-114$

As the remainder is -114, the term x+2 is not a factor of given polynomial

The correct answer is: No, the remainder is -114

Keywords: Polynomials, expressions

Learn more about polynomials at:

#LearnwithBrainly

3 0

3 years ago

Based on the given information, what is the measure of the missing length, c?

salantis [7]

Answer:

$c=5\ units$

Step-by-step explanation:

we know that

In the right triangle ABC of the figure

Applying the Pythagoras Theorem

$AC^{2}=AB^{2}+BC^{2}$

substitute the given values

$13^{2}=c^{2}+12^{2}$

Solve for c

$169=c^{2}+144$

$c^{2}=169-144$

$c^{2}=25$

$c=5\ units$

4 0

3 years ago

A simulation was conducted using 10 fair six-sided dice, where the faces were numbered 1 through 6. respectively. All 10 dice we

kompoz [17]

Answer:

C) a sample distribution of a sample mean with n = 10

$\mu_{{\overline}{X}} = 3.5$

and $\sigma_{{\overline}{Y}} = 0.38$

Step-by-step explanation:

Here, the random experiment is rolling 10, 6 faced (with faces numbered from 1 to 6) fair dice and recording the average of the numbers which comes up and the experiment is repeated 20 times.So, here sample size, n = 20 .

Let,

$X_{ij}$ = The number which comes up on the ith die on the jth trial.

∀ i = 1(1)10 and j = 1(1)20

Then,

$E(X_{ij})$ = $\frac {1 + 2 + 3 + 4 + 5 + 6}{6}$

= 3.5 ∀ i = 1(1)10 and j = 1(1)20

and,

$E(X^{2}_{ij}$ = $\frac {1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}}{6}$

= $\frac {1 + 4 + 9 + 16 + 25 + 36}{6}$

= $\frac {91}{6}$

$\simeq$ 15.166667

so, $Var(X_{ij}$ = $(E(X^{2}_{ij} - {(E(X_{ij})}^{2})$

$\simeq 15.166667 - 3.5^{2}$

= 2.91667

and $\sigma_{X_{ij}}$ = $\sqrt {2.91667}[/tex [tex]\simeq 1.7078261036$

Now we get that,

$Y_{j} = \frac {\sum_{j = 1}^{20}X_{ij}}{20}$

We get that $Y_{j}'s$ are iid RV's ∀ j = 1(1)20

Let, ${\overline}{Y} = \frac {\sum_{j = 1}^{20}Y_{j}}{20}$

So, we get that $E({\overline}{Y}) = E(Y_{j})$

= $E(X_{ij}$ for any i = 1(1)10

= 3.5

and,

$\sigma_{({\overline}{Y})} = \frac {\sigma_{Y_{j}}}{\sqrt {20}} = \frac {\sigma_{X_{ij}}}{\sqrt {20}} = \frac {1.7078261036}{\sqrt {20}} [tex]\simeq 0.38$

Hence, the option which best describes the distribution being simulated is given by,

C) a sample distribution of a sample mean with n = 10

$\mu_{{\overline}{X}} = 3.5$

and $\sigma_{{\overline}{Y}} = 0.38$

6 0

2 years ago

What is the solution to (x + 8) > (2x + 10)?<br> (–∞, –4)<br> (–4, ∞)<br> (–∞, 4)<br> (4, ∞)

Citrus2011 [14]

Answer:

(–∞, -4)

Step-by-step explanation:

5 0

2 years ago