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3 years ago

Write the equation in point-slope form

Mathematics

1 answer:

Reptile [31]3 years ago

8 0

Answer:

first i know but idk then

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. The cost of 2 pizzas and 1 burger is ₹ 450. Write a linear equation for this situation and draw its graph.

Akimi4 [234]

Answer:

The linear equation is;

Y = 450 - 2·X

Please find the included graph

Step-by-step explanation:

Whereby we have the following relation;

The cost of 1 pizza = X

The cost of 1 burger = Y

Hence;

450 = Y + 2·X

Which gives;

Y = 450 - 2·X

The linear equation for the situation is therefore as presented above

The graph of the linear equation can be plotted using the assumed data as follows;

Y, X

1, 448

2, 446

3, 444

4, 442

5, 440

6, 438

7, 436

8, 434

9, 432

10, 430

11, 428

12, 426

13, 424

14, 422

15, 420

16, 418

8 0

3 years ago

A ÷ (2 + 1.5) for a = 14<br><br> dont spam links and this is 6th grade work

IRINA_888 [86]

Answer:

Step-by-step explanation:

2+1.5=3.5

49/3.5=14

4 0

2 years ago

Read 2 more answers

Explain on the next slide how in 3 hours the car wouldn't travel 160 miles.

rosijanka [135]

Answer:

the car would have to be going under 5mph

Step-by-step explanation:

5 0

3 years ago

In a survey of 225 art students 28% reported that their favorite color is blue. What is the margin of error for the sample?

ipn [44]

Margin of error is important in determining the probability as it gives realistic data for the distribution. The formula to be followed for margin of error is m% = z* sqrt (p*(1-p))/n where p is the % positive outcome, n is the sample size and z* is a constant dependent to % confidence level. Substituting, m% = z* (0.03). Level of confidence is not given but assuming at95% CF, m% = 6%

3 0

3 years ago

Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean we

Marina86 [1]

Answer:

Required Probability = 0.1283 .

Step-by-step explanation:

We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.

Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.

Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.

Let X = mean weight of the babies, so X ~ N( $\mu = 7 lbs , \sigma^{2} = 0.875^{2} lbs$ )

The standard normal z distribution is given by;

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

where, X bar = sample mean weight

n = sample size = 4

Now, probability that the average weight of the four babies will be more than 7.5 lbs = P(X bar > 7.5 lbs)

P(X bar > 7.5) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{7.5-7}{\frac{0.875}{\sqrt{4} } }$ ) = P(Z > 1.1428) = 0.1283 (using z% table)

Therefore, the probability that the average weight of the four babies will be more than 7.5 lbs is 0.1283 .

8 0

4 years ago