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2 years ago

PLS I NEED HELP I’LL MARK BRAINLESS

Mathematics

2 answers:

vfiekz [6]2 years ago

4 0

Answer: 17, 13, 9, 5, 1

Step-by-step explanation:

y=-2(-2) + 9 = 13

y=-2(0) + 9 = 9

y=-2(2) + 9 = 5

y=-2(4) + 9 = 1

dmitriy555 [2]2 years ago

3 0

You plug in the numbers for x and solve the inequality for y.

These should be the equations in order from left to right.

Hope this helps and good luck!

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Katy chose a card at random from a deck of 52 cards. Katy replaced the first card and then chose a second card. What is the prob

Alisiya [41]

52 total cards....13 total diamonds in the deck

P(1st card picked being a diamond) = 13/52
replacing the card, so the sample space remains the same
P(2nd card being a diamond) = 13/52

P(both) = 13/52 * 13/52 = 169/2704 = 1/16 <==

5 0

3 years ago

Read 2 more answers

What is the charge for 400 text messages? A. $50 B. $ 180 C. $90 D.$100

nadezda [96]

Cost = .10 x ( n ) + 50
= .10(400) + 50
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I hope this helps.

6 0

2 years ago

Read 2 more answers

2. CTfastrak bus waiting times are uniformly distributed from zero to 20 minutes. Find the probability that a randomly selected

Juliette [100K]

Answer:

b. 0.25

c. 0.05

d. 0.05

e. 0.25

Step-by-step explanation:

if the waiting time x follows a uniformly distribution from zero to 20, the probability that a passenger waits exactly x minutes P(x) can be calculated as:

$P(x)=\frac{1}{b-a}=\frac{1}{20-0} =0.05$

Where a and b are the limits of the distribution and x is a value between a and b. Additionally the probability that a passenger waits x minutes or less P(X<x) is equal to:

$P(X$

Then, the probability that a randomly selected passenger will wait:

b. Between 5 and 10 minutes.

$P(5$

c. Exactly 7.5922 minutes

$P(7.5922)=0.05$

d. Exactly 5 minutes

$P(5)=0.05$

e. Between 15 and 25 minutes, taking into account that 25 is bigger than 20, the probability that a passenger will wait between 15 and 25 minutes is equal to the probability that a passenger will wait between 15 and 20 minutes. So:

$P(15$