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3 years ago

25 POINTS! Please help me with these two questions. Ive been stuck on them for over an hour. :(

Mathematics

1 answer:

Ber [7]3 years ago

8 0

Answer:

m=1/4

y=4

b=0

This should be the answer

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Plz Help

Afina-wow [57]

We are asked in the problem to determine the value of x in the equation 4x² + 2x = 7 using completing the square method. in this case,

4x² + 2x = 7
x2 + 0.5 = 1.75
(x + 0.25)^2 - 0.0625 = 1.75
(x + 0.25)^2 = 1.8125
x+0.25=1.3462
x1 =1.0963
x2 =-1.5962

3 0

3 years ago

How much is 5/9 of -3/5? A) 1/2 B) -1/3 C) 1/7 D) -25/27

irina1246 [14]

The answer is A 1/2 sorry if I’m wrong if not A its c

5 0

3 years ago

Read 2 more answers

In the morning Mary walked 2 1/4 miles around the park. In the afternoon she walked another 5 7/8. how many miles in total did s

anygoal [31]

Answer:

8 1/8 total miles walked.

Step-by-step explanation:

I like to make fractions simple so I would do this,

9/4 plus 47/8

Take 8 and your common denominator and solve.

18/8 plus 47/8 = 65/8 which can be put as 8 1/8.

7 0

3 years ago

What is the answer to 3x + 2 = -19?

Marrrta [24]

Answer:

X = -7

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

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

3 years ago