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2 years ago

NEED HELP ASAP PLEASE!

Mathematics

2 answers:

Ivan2 years ago

5 0

Answer:

i think the answer is b

Step-by-step explanation:

11111nata11111 [884]2 years ago

5 0

Answer:

The answer is y=3x+2

It's the third option

Step-by-step explanation:

The formula for slope is y= mX+b

m= the slope

b= the y intercept

Since the y intercept is 2 that means b= 2

To find the slope or "m" you would find a point on the graph and find the rate of change to the y intercept wich is 3.

Plug them all in and you get y= 3x+2

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Solve the equation −96=3(8x)^(5/3).

vladimir2022 [97]

Answer:

x= - 1

Step-by-step explanation:

5 0

2 years ago

Please help me I'll give you a brainliest

astraxan [27]

Okay try to round the whole number and see if you can solve that should help you

7 0

2 years ago

If mc022-1.j pg and mc022-2.j pg, what is the value of (f – g)(144)? –84 –60 0 48

Mashutka [201]

Answer:

Step-by-step explanation:

f(x) = √(x) + 12

g(x) = 2√(x)

(f-g)(x) = √(x) + 12 - 2√(x)

(f-g)(x) = 12 - √(x)

if x = 144

(f-g)(144) = 12 - √(144) = 12 - 12 = 0

5 0

3 years ago

Read 2 more answers

the value y varies directly with x. which function represents the relationship between x and y if y=20/3 when x=30?

joja [24]

If y varies directly with x, means that they can be modeled by a linear equation, lets choose a line with 0 y intercept, that is:
y = mx + b
y = mx
where m is the slope of the line, now we plug in the data we have:
y = mx
20/3 = m(30)
solving for m:
m = (20/3)(1/30)
m = 20/90
m = 2/9
so the line equation, or function modeling the y and x relationship is:
y = (2/9)x

6 0

2 years ago

A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean $\mu$ and standard deviation $\sigma$ , we have these following probabilities

$Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827$

$Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545$

$Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973$

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So $\mu = 53, \sigma = 11$

So:

$Pr(53-11 \leq X \leq 53+11) = 0.6827$

$Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545$

$Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973$

-----------

$Pr(42 \leq X \leq 64) = 0.6827$

$Pr(31 \leq X \leq 75) = 0.9545$

$Pr(20 \leq X \leq 86) = 0.9973$

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have $Pr(31 \leq X \leq 75) = 0.9545 = 0.9545$ subtracted by $Pr(42 \leq X \leq 64) = 0.6827$ is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.

To find just the percentage above the mean, we divide this value by 2

So:

$P = {0.9545 - 0.6827}{2} = 0.1359$

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

4 0

3 years ago