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andrew-mc [135]
4 years ago
6

Tessie has a volleyball game at 6:45 p.m. she needs to be there 15 minutes early to warm up for the game and it takes her 40 min

utes to go to The gym what time should she leave her house
Mathematics
1 answer:
const2013 [10]4 years ago
6 0
First we need to add the time she takes warming and the time she takes going to the the gym:
15minutes+40minutes=45minutes

Next, we area going to subtract those 45 minutes from the time the game begins: 
6:45pm-45minutes=6:00pm

We can conclude that she should leaver her house at 6:00 pm if she wants to get in time for her 6:45 pm game.
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A. one solution <br> B. infinitely many solutions <br> C. no solution
kodGreya [7K]
It should be A-one solution
3 0
2 years ago
According to a​ survey, 65​% of murders committed last year were cleared by arrest or exceptional means. Fifty murders committed
monitta

Answer:

a) P(X=41)=(50C41)(0.65)^{41} (1-0.65)^{50-41}=0.00421

b) P(X=36)=(50C36)(0.65)^{36} (1-0.65)^{50-36}=0.0714

P(X=37)=(50C37)(0.65)^{37} (1-0.65)^{50-37}=0.0502

P(X=38)=(50C38)(0.65)^{38} (1-0.65)^{50-38}=0.0319

And adding these values we got:

P(36 \leq X \leq 38)= 0.1535

c) We can find the expected value given by:

E(X) = np =50*0.65 = 32.5

And the standard deviation would be:

\sigma = \sqrt{np(1-p)} \sqrt{50*0.65*(1-0.65)}= 3.373

We can use the approximation to the normal distribution and we have at leat 95% of the data within 2 deviations from the mean. And the lower limit for this case would be:

\mu -2\sigma = 32.5- 2*3.373 = 25.75

And then we can consider a value of 18 as unusual lower for this case.

Step-by-step explanation:

Let X the random variable of interest "number cleared by arrest or exceptional", on this case we can model this variable with this distribution:

X \sim Binom(n=50, p=0.65)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Part a

We want this probability:

P(X=41)=(50C41)(0.65)^{41} (1-0.65)^{50-41}=0.00421

Part b

We want this probability:

P(36 \leq X \leq 38)

We can find the individual probabilities:

P(X=36)=(50C36)(0.65)^{36} (1-0.65)^{50-36}=0.0714

P(X=37)=(50C37)(0.65)^{37} (1-0.65)^{50-37}=0.0502

P(X=38)=(50C38)(0.65)^{38} (1-0.65)^{50-38}=0.0319

And adding these values we got:

P(36 \leq X \leq 38)= 0.1535

Part c

We can find the expected value given by:

E(X) = np =50*0.65 = 32.5

And the standard deviation would be:

\sigma = \sqrt{np(1-p)} \sqrt{50*0.65*(1-0.65)}= 3.373

We can use the approximation to the normal distribution and we have at leat 95% of the data within 2 deviations from the mean. And the lower limit for this case would be:

\mu -2\sigma = 32.5- 2*3.373 = 25.75

And then we can consider a value of 18 as unusual lower for this case.

6 0
3 years ago
Can someone please help meeee ?
nataly862011 [7]
It could possibly be the first on but I’m not completely sure
5 0
3 years ago
What is the solution to the equation
Dafna11 [192]
\sqrt{4t+5}=3-\sqrt{t+5}\\\\D:4t+5\geq0\ \wedge\ t+5\geq0\ \wedge\ \sqrt{t+5}\leq3\\\\t\geq-\dfrac{5}{4}\ \wedge\ t\geq-5\ \wedge\ t\leq6

therefore
D:x\in\left< -\dfrac{5}{4};\ 6\right>

\sqrt{4t+5}=3-\sqrt{t+5}\ \ \ |^2\\\\(\sqrt{4t+5})^2=(3-\sqrt{t+5})^2\ \ \ |use:(a-b)^2=a^2-2ab+b^2\\\\4t+5=3^2-2\cdot3\cdot\sqrt{t+5}+(\sqrt{t+5})^2\\\\4t+5=9-6\sqrt{t+5}+t+5\ \ \ \ |-t\\\\3t+5=14-6\sqrt{t+5}\ \ \ \ |-14\\\\3t-9=-6\sqrt{t+5}\ \ \ \ |change\ signs
9-3t=6\sqrt{t+5}\ \ \ \ |:3\\\\3-t=2\sqrt{t+5}\ \ \ \ |^2\\\\(3-t)^2=(2\sqrt{t+5})^2\\\\3^2-2\cdot3\cdot t+t^2=4(t+5)\\\\9-6t+t^2=4t+20\ \ \ |-4t-20\\\\t^2-10t-11=0\\\\t^2+t-11t-11=0\\\\t(t+1)-11(t+1)=0\\\\(t+1)(t-11)=0\iff t+1=0\ \vee\ t-11=0\\\\t=-1\in D\ \vee\ t=11\notin D
Answer: t = -1.


7 0
3 years ago
Read 2 more answers
Help a bab out with something, please?
Nuetrik [128]

Answer:

18

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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