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1 year ago

Help me! ill give brainiest. this is my finals and im not doing good!

Mathematics

2 answers:

vredina [299]1 year ago

8 0

1 3/9 is the answer your looking for

Alex73 [517]1 year ago

4 0

Answer:

question 1 = 1 3/9 question 2 = 8/9 question 3 = 11/22

Step-by-step explanation:

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These points are linear.<br> Find the slope.<br> x0 |1 |2|3|4|5<br> y|0|2|4|6|8|10<br> slope = [?]

jolli1 [7]

Answer:

Step-by-step explanation:

x₁ = 0 ; y₁=0 &

x₂ = 1 ; y₂=2

Slope = $\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$=\dfrac{2-0}{1-0}\\\\\\=\dfrac{2}{1}\\\\\\= 2$

Slope = 2

7 0

2 years ago

Select each expression that is equivalent to 3n+6 select all that apply. Explan.

arsen [322]

A, C. A is obviously correct because it’s the e same thing and C you combine like terms (2n+n=3n,2+4=6,=3n+6) it’s not B because it’s not 18, it’s not D because you’d get 3n+18 and it’s not E because you’d get 3n+12, meaning the only correct answers are A and C

5 0

2 years ago

The expected number of typographical errors on a page of a certain magazine is .2. What is the probability that an article of 10

Pavel [41]

Answer:

a) The probability that an article of 10 pages contains 0 typographical errors is 0.8187.

b) The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.

Step-by-step explanation:

Given : The expected number of typographical errors on a page of a certain magazine is 0.2.

To find : What is the probability that an article of 10 pages contains

(a) 0 and (b) 2 or more typographical errors?

Solution :

Applying Poisson distribution,

$N\sim Pois(0.2)$

$P(N=r)=\frac{e^{-np}(np)^r}{r!}$

where, n is the number of words in a page

and p is the probability of every word with typographical errors.

Here, n=10 and E(N)=np=0.2

a) The probability that an article of 10 pages contains 0 typographical errors.

Substitute r=0 in formula,

$P(N=0)=\frac{e^{-0.2}(0.2)^0}{0!}$

$P(N=0)=\frac{e^{-0.2}}{1}$

$P(N=0)=e^{-0.2}$

$P(N=0)=0.8187$

The probability that an article of 10 pages contains 0 typographical errors is 0.8187.

b) The probability that an article of 10 pages contains 2 or more typographical errors.

Substitute $r\geq 2$ in formula,

$P(N\geq 2)=1-P(N$

$P(N\geq 2)=1-[P(N=0)+P(N=1)]$

$P(N\geq 2)=1-[\frac{e^{-0.2}(0.2)^0}{0!}+\frac{e^{-0.2}(0.2)^1}{1!}]$

$P(N\geq 2)=1-[e^{-0.2}+e^{-0.2}(0.2)]$

$P(N\geq 2)=1-[0.8187+0.1637]$

$P(N\geq 2)=1-0.9825$

$P(N\geq 2)=0.0175$

The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.

6 0

2 years ago

From a group of 3 men and 4 women, a delegation of 2 is selected. What is the expected number of men in the delegation?

Ostrovityanka [42]

Answer:

The expected number of men in the delegation is 0.86.

Step-by-step explanation:

Number of men = 3

Number of women = 4

A delegation of 2 is selected.

Let X be the number of men selected. So, the possible values of X are 0,1,2.

$P(X=0)=\dfrac{^3C_0\times ^4C_2}{^7C_2}=\dfrac{2}{7}$

$P(X=1)=\dfrac{^3C_1\times ^4C_1}{^7C_2}=\dfrac{4}{7}$

$P(X=2)=\dfrac{^3C_2\times ^4C_0}{^7C_2}=\dfrac{1}{7}$

The expected number of men in the delegation is

$E(x)=0\times P(X=0)+1\times P(X=1)+2\times P(X=2)$

$E(x)=0\times \dfrac{2}{7}+1\times \dfrac{4}{7}+2\times \dfrac{1}{7}$

$E(x)=\dfrac{6}{7}$

$E(x)\approx 0.86$

Therefore, the expected number of men in the delegation is 0.86.

4 0

2 years ago

Let g(x) = 2x and h(x) = x2 + 4. Evaluate (h ∘ g)(−5).

Zolol [24]

Hi! the answer should be 104

----

Formula: h(g(x))

1) First you want to take the -5 and plug that into the given equation for g(x)

• 2(-5)= -10

2) Next you want to take the -10 and plug it into the equation for h(x)

• (-10)²+4= 104

6 0

1 year ago