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2 years ago

PLEASE HELP WILL GIVE BRAINLIEST!!!!! A survey asked students whether they have any siblings and pets.

Mathematics

2 answers:

LiRa [457]2 years ago

7 0

The answrr is B, 60%. 45/75 equals 60, so it should be 60%.

siniylev [52]2 years ago

3 0

Answer:

Option B.

Step-by-step explanation:

The given table represents the relative frequency.

Let A and B represent the following events.

A : Student does not have a sibling

B : Students has a pet.

Then intersection of A and B (A∩B) denotes the event that the student does not have a sibling but has a pet.

Using the given table it is clear that

P(A∩B) = 0.15

We need to find the probability that the student does not have a pet given that student has a sibling. It means we need to find the value of P(B|A).

Formula for conditional probability:

$P(B|A)=\dfrac{P(A\cap B)}{P(A)}$

Substitute the given values.

$P(B|A)=\dfrac{0.15}{0.25}$

$P(B|A)=\dfrac{3}{5}$

$P(B|A)=0.6=60\%$

The probability that the student does not have a pet given that student has a sibling is 60%.

Therefore, the correct option is B.

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A triangle whose angles have measures x x+5 and x+10

nirvana33 [79]

the correct question is
A triangle whose angles have measures x x+5 and x+10.How do you find the measure of each angle?

we know that
The Sum of internal angles of any triangle = 180∘
so
[x+(x+5)+(x+10)]=180---------------> 3x+15=180----------> 3x=165-------> x=55°

the answer is
the measure of each angle is
x=55°
x+5=60°
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3 years ago

7(2 + 3) − 4 · 2 2 the second two is up in the air fyi

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Step-by-step explanation:

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2 years ago

Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E

Ipatiy [6.2K]

Answer:

(A) $y=ke^{2t}$ with $k\in\mathbb{R}$ .

(B) $y=ke^{2t}/2-1/2$ with $k\in\mathbb{R}$

(D) $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ ,

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then $0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}$ . With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form $e^{2t}$ with $t$ real.

(B) Proceeding and the previous item, we obtain $1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2$ . Which is not a vector space with the usual operations (this is because $-1/2$ ), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set $y=e^{mt} \Rightarrow y''=m^2e^{mt}$ and we obtain the characteristic equation $0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$ , and as in the first items the set of solutions form a vector space.

(D) Using C, let be $y=me^{3t}$ we obtain that it must satisfies $3^2m-4m=1\Rightarrow m=1/5$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ , and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).

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3 years ago

Round 478.54 to the nearest tenth

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If the number after the tenth if four or lower, then the tenth stays the same value as it is
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