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2 years ago

PLEASE SOMEONE DO THIS FOR ME: Determine the x - and y - intercepts of the following linear function: 4x + 8y =16

Mathematics

2 answers:

ra1l [238]2 years ago

8 0

4x+8y=16

4x/4+8y/8=16÷4/8

x+y=16×8/4

x+y=16×2

x+y=32

Kitty [74]2 years ago

3 0

Answer:

x = 2 and y = 1

Step-by-step explanation:

4*2 = 8

8 * 1 =8

8+8=16

4(2) + 8(1) = 16

could pls have brainliest!!

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What is the amswer to -4k-2(k-3)=-6k+8

SashulF [63]

Answer:

k = 1/2

Step-by-step explanation:

-4k-2(k)-2(-3)=-6k+8

-4k-2k+6=-6k+8

-2k+6=-6k+8

4k+6=8

4k=2

4k/4= 2/4

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3 years ago

After a discount of 15%, the of a shirt is $51. what was the original price of the shirt?

lina2011 [118]

Answer:

$58.65

Step-by-step explanation:

change 15% to a decimal than multiply .15 by 51 then add 51.

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3 years ago

Read 2 more answers

1 and 16/24 + 1 and 21/24

AlexFokin [52]

Answer:

turn into improper fractions

1*24+16=40

40/24

1*24+21=45

45/24

40/24+45/24=85/24

work out how many times 24 goes into 85 to get back to a mixed number.

24 goes into 85 three times with 13 left over so the answer is

3 and 13/24

6 0

2 years ago

Write down 6 numbers that have a median of 8, a mean of 9 and a range of 13

spayn [35]

Answer:

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Step-by-step explanation:

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3 years ago

The service department of a luxury car dealership conducted research on the amount of time its service technicians spend on each

mart [117]

Answer:

Probability that the mean service time is between 1 and 2 hours is 0.96764.

Step-by-step explanation:

We are given that a systematic random sample of 100 service appointments has been collected.

The 100 appointments showed an average preparation time of 90 minutes with a standard deviation of 140 minutes.

Let $\bar X$ = sample mean service time

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = average preparation time = 90 minutes

$\sigma$ = standard deviation = 140 minutes

n = sample of appointments = 100

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, probability that the mean service time is between 60 and 120 minutes is given by = P(60 minutes < $\bar X$ < 120 minutes)

P(60 minutes < $\bar X$ < 120 minutes) = P( $\bar X$ < 120 min) - P( $\bar X$ $\leq$ 60 min)

P( $\bar X$ < 120 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{120-90}{\frac{140}{\sqrt{100} } }$ ) = P(Z < 2.14) = 0.98382

P( $\bar X$ $\leq$ 60 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{60-90}{\frac{140}{\sqrt{100} } }$ ) = P(Z $\leq$ -2.14) = 1 - P(Z < 2.14)

= 1 - 0.98382 = 0.01618

The above probability is calculated by looking at the value of x = 2.14 in the z table which has an area of 0.98382.

Therefore, P(60 min < $\bar X$ < 120 min) = 0.98382 - 0.01618 = 0.96764

7 0

3 years ago