MATHEMATICS COLLEGE

Help me three questions please really easy I promise answer

Answers

Answer 1

Answer:

Answer:

1.y=-2x+7

2. no

3.yes

Step-by-step explanation:

Related Questions

Please help me with this
The computer that controls a bank's automatic teller machine crashes a mean of 0.5 times per day. What is the probability that, in any seven-day week, the computer will crash less than 3 times? Round your answer to four decimal places.
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Help asap. yes i will give brainliest

Answers

Answer:

A = 1.5
B = -1/3
C = -4/3

Step-by-step explanation:

For A:

A is smack in the middle of 1 and 2, so it it 1 1/2 or 1.5 (same thing)

For B:

Each tick mark between the numbers represents 1/3
B is on the first tick mark between 0 and -1, so therefore it is -1/3

For C:

C is on the first tick mark between -1 and -2
That means C is -1 1/3
-1 1/3 = -4/3

I hope this helps!

Answer:

A ----> +1.5

B------> -0.33

C-------> -0.1322

Step-by-step explanation:

$A\ is\ +1.5\ as\ it\ is\ 1.5\ units\ upwards(positively)\ from\ 0.\nB\ is\ -(1)/(3)\ as\ it\ equates\ to\ -0.33\ which\ is\ between\ 0\ and\ -1.\nC\ is\ -(4)/(3)\ as\ it\ equates\ to\ -1.322\ which\ is\ between\ -1\ and\ -2.\n$

A survey of cars on a certain stretch of highway during morning commute hours showed that 70% had only one occupant, 15% had 2, 10% had 3, 3% had 4, and 2% had 5. Let Xrepresent the number of occupants in a randomly chosen car.a. Find the probability mass function of X.

b. Find P(X ≤ 2).

c. Find P(X > 3).

d. Find μX.

e. Find σX

Answers

Answer:

a) X 1 2 3 4 5

P(X) 0.7 0.15 0.10 0.03 0.02

b) $P(X \leq 2) = P(X=1) +P(X=2) = 0.7+0.15=0.85$

c) $P(X >3) = 1-P(X \leq 3) = 1-[P(X=1) +P(X=2)+P(X=3)]=1-[0.7+0.15+0.1]= 0.05$

d) $E(X) = \sum_(i=1)^n X_i P(X_i) = 1*0.7 +2*0.15+ 3*0.1+4*0.03+ 5*0.02= 1.52$

e) $E(X^2) = \sum_(i=1)^n X^2_i P(X_i) = 1*0.7 +4*0.15+ 9*0.1+16*0.03+ 25*0.02=3.18$

$Var(X) = E(X^2) -[E(X)]^2= 3.18- (1.52)^2 = 0.8996$

$\sigma= √(Var(X))= √(0.8996)= 0.933$

Step-by-step explanation:

Part a

From the information given we define the probability distribution like this:

X 1 2 3 4 5

P(X) 0.7 0.15 0.10 0.03 0.02

And we see that the sum of the probabilities is 1 so then we have a probability distribution

Part b

We want to find this probability:

$P(X \leq 2) = P(X=1) +P(X=2) = 0.7+0.15=0.85$

Part c

We want to find this probability $P(X>3)$

And for this case we can use the complement rule and we got:

$P(X >3) = 1-P(X \leq 3) = 1-[P(X=1) +P(X=2)+P(X=3)]=1-[0.7+0.15+0.1]= 0.05$

Part d

We can find the expected value with this formula:

$E(X) = \sum_(i=1)^n X_i P(X_i) = 1*0.7 +2*0.15+ 3*0.1+4*0.03+ 5*0.02= 1.52$

Part e

For this case we need to find first the second moment given by:

$E(X^2) = \sum_(i=1)^n X^2_i P(X_i) = 1*0.7 +4*0.15+ 9*0.1+16*0.03+ 25*0.02=3.18$

And we can find the variance with the following formula:

$Var(X) = E(X^2) -[E(X)]^2= 3.18- (1.52)^2 = 0.8996$

And we can find the deviation taking the square root of the variance:

$\sigma= √(Var(X))= √(0.8996)= 0.933$

Each prize for a carnival booth costs $0.32. How many prizes can you buy with $96?

Answers

Answer: 96 divided by 0.32 is 300, so it's 300.

An attacker at the base of a castle wall 4 m high throws a rock straight up with speed 7.5 m/s from a height of 1.5 mabove the ground.
(a) Will the rock exceed the top of the wall?
(b) If so, what is its speed when it reaches the top of the wall?
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(c) If we don’t want the stone to exceed the top of the wall (we want the top of the wall to be the maximum height)
what the initial speed that the stone must have?