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2 years ago

Please help!! I’ll give brainliest

Mathematics

1 answer:

Tatiana [17]2 years ago

3 0

Answer:

4 and 2/3 from choice A on the promblem

You might be interested in

I am having trouble with this problem. I need to find the lengths of this right triangle. How would you do this?

podryga [215]

You labeled the triangle wrong sides 'a' and 'b' are supposed to be the sides that make the right angle. the other side is called the hypotenuse which is the longest side which you should have labeled 'c'

so Pythagorean theorem says
a^2+b^2=c^2
so
(2x+1)^2+(11x+5)^2=(12x+1)^2
distribute
(4x^2+4x+1)+(121x^2+110x+25)=(144x^2+24x+1)
add like terms
125x^2+114x+26=144x^2+24x+1
subtract 125x^2 from both sides
114x+26=19x^2+24x+1
subtract 114x from both sides
26=19x^2-90x+1
subtract 26 from both sides
0=19x^2-90-25
factor
(x-5)(19x+5)=0
therefor x-5=0 and/or 19x+5=0
so
x-5=0 add 5 to both sides
x=5
19x+5=0
subtract 5 from both sides
19x=-5
divide both sides by 19
x=-5/19
since side legnths can't be negative, we can cross this solution out

so x=5
subtitute
1+2x
1+2(5)
1+10=11
side a=11

11x+5
11(5)+5
55+5=60
side b=60

12x+1
12(5)+1
60+1=60
side c=61
add them all up
side a+b+c=11+60+61=132=total legnth

5 0

3 years ago

The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(

Umnica [9.8K]

Answer:

(a) $A_1$ and $A_2$ are indeed mutually-exclusive.

(b) $\displaystyle P(A_1\; \cap \; B) = \frac{1}{20}$ , whereas $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

(d) $\displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}$ , whereas $P(A_1 \; |\; B) = \displaystyle \frac{4}{9}$

Step-by-step explanation:

$P(A_1 \; \cap \; A_2) = 0$ means that it is impossible for events $A_1$ and $A_2$ to happen at the same time. Therefore, event $A_1$ and $A_2$ are mutually-exclusive.

By the definition of conditional probability:

$\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}$ .

Rearrange to obtain:

$\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot P(A_1) = 0.25 \times 0.20 = \frac{1}{20}$ .

Similarly:

$\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot P(A_2) = 0.80 \times 0.05 = \frac{1}{25}$ .

Note that:

$\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}$ .

In other words, $A_1$ and $A_2$ are collectively-exhaustive. Since $A_1$ and $A_2$ are collectively-exhaustive and mutually-exclusive at the same time:

$\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}$ .

By Bayes' Theorem:

$\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}$ .

Similarly:

$\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}$ .

6 0

3 years ago

A survey of 35 people was conducted to compare their self-reported height to their actual height. The difference between reporte

Over [174]

Answer:

The test statistic is t = 3.36.

Step-by-step explanation:

You're testing the claim that the mean difference is greater than 0.7.

At the null hypothesis, we test if the mean difference is of 0.7 or less, that is:

$H_0: \mu \leq 0.7$

At the alternate hypothesis, we test if the mean difference is greater than 0.7, that is:

$H_1: \mu > 0.7$

The test statistic is:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, s is the standard deviation of the sample and n is the size of the sample.