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Mrac [35]
3 years ago
6

Anthony will be competing in a body building competition in 12 months. He initially weighs 160 pounds and by the end of 12 month

s weighs 232 pounds.
What is the approximate rate of change in Anthony's body weight?
A) 6 pounds per month
B) 13 pounds per month
C) 19 pounds per month
D) 36 pounds per month
Mathematics
2 answers:
aksik [14]3 years ago
3 0
A) 6 pounds per month
grigory [225]3 years ago
3 0
To find the rate<span> of change, find the total change in weight and then divide it by number of months over which it changed. So the answer is </span><span>72/12</span><span> which is </span>6 pounds per month<span>. so the answer is (A.6 pounds per month</span>
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kolezko [41]

Answer:

I use app called math

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Step-by-step explanation:

7 0
2 years ago
PLZZZZZZZZZZ I WILL GIVE BRAINLIEST AND 15PTS!!!!!!!!!!!!!!
vazorg [7]

Answer: ( 0, 2 )

Step-by-step explanation:

You have to put coordinates of each graph on these equations y = -x + 2 and y = (1/2)x + 2. If putting coordinates satisfies both equations, then that coordinate will be the solution.

For example, let's put (0, 2) to equations.

y = -x + 2

2 = -0 + 2

2 = 2, true

y = (1/2)x + 2

2 = (1/2) × 0 + 2

2 = 2, true

So, ( 0, 2 ) is the solution.

7 0
3 years ago
Read 2 more answers
The price of a car has been reduced from $14,000 to $8,400. What is the percentage decrease of the price of the car?
Lady bird [3.3K]

Answer: 40%

Step-by-step explanation:

If you divide 8400 by 14000. You will see that $8400 is 60% (.60 in the calculator) you need to subtract that from 100 to find the discount. 100-60= 40

3 0
3 years ago
On average, Tanner has noticed that 22 trains pass by his house daily (24 hours) on the nearby train tracks. What is the probabi
blondinia [14]

Answer:

0.11069

Step-by-step explanation:

We will assume that the trains pass by his house following a uniform distribution with values between 0 and 24. The probability of a train passing on a 9-hour time period is 9/24 = 3/8 = 0.375. Lets call Y the amount of trains passing by his house during that 9-hour period. Y follows a Binomail distribution with parameters 22 and 0.375.

P(Y ≤ 5) = P(Y = 0) + P(Y=1) + P(Y=2) + P(Y=3) + P(Y=4) + P(Y=5) =

{22 \choose 0} * 0.375^0*(1-0.375)^{22} + {22 \choose 1}*0.375^1*(1-0.375)^{21} +\\\\{22 \choose 2} * 0.375^2*(1-0.375)^{20} + {22 \choose 3}*0.375^3*(1-0.375)^{19}  + \\{22 \choose 4} * 0.375^4*(1-0.375)^{18} + {22 \choose 5}*0.375^5*(1-0.375)^{17} = 0.11069

I hope that works for you!

4 0
3 years ago
The probability that a customer of a network operator has a problem about you needing technical staff's help in a month is 0.01.
snow_tiger [21]

Answer:

(a) average calls = 5  

(b) probability that there is exactly one call in 6 consecutive monts = 0.038

Step-by-step explanation:

Let event of a customer requiring help in a particular month = H

and event of a customer not requiring help in a particular month = ~H

Given

p= 0.01,  therefore

Number of households, n = 500.

Binomial distribution:

x = number of households requiring help in a particular month

P(x,n,p) = C(x,n)*p^x*(1-p)^(n-x)

where

C(x,n) = n!/(x!(n-x)!) is the the number of combinations of x objects out of n

(a) Average number of households requiring help = np = 500*0.01 = 5

(b)

Probability that there are no calls requiring help in a particular month

P(0), q= C(0,n)*p^0(1-p)^(n-0)

= 1*1*0.99^500

= 0.006570483

Applying binomial probability over six months,

q = 0.006570483

n = 6

x = 1

P(x,n,q)

= C(x,n)*q^x*(1-q)^(n-x)

= 6!/(1!*5!) * 0.006570483^1 * (1-0.006570483)^5

= 0.038145

Therefore the probability that in 6 consecutive months there is exactly one month that no customer has called for help = 0.038

3 0
2 years ago
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