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3 years ago

Graph the equation. y = 1/3 (x + 3)(x – 3)

Mathematics

2 answers:

kolbaska11 [484]3 years ago

6 0

Answer:

Step-by-step explanation:

screen shot attached

love history [14]3 years ago

5 0

I did the work and I screenshot it

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Use the net to compute the surface area of the three-dimensional figure. A) 210 units B) 225 units C) 240 units D) 280 units

11111nata11111 [884]

The answer would most commonly be C only because a square is 360 degrees and the surface area would be a 6th of that which is 60 then i multiplied it by 4 because of the four corners signaling the height and or length of the 2D figure square which gives me C.

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2 years ago

( 1/4) ꓥ -3 = A= -64 B= -1/64 C= 1/64 D= 64

Anna11 [10]

Answer:

C-1/64

Step-by-step explanation:

(1/4)(1/4)(1/4)=1/64

Mark the brainliest

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2 years ago

Step by step what is 3 (x+2)+5x

lesya692 [45]

Answer:

8x+6

Step-by-step explanation:

To simplify it, the first thing you want to do is distribute the three.

3(x+2) = 3x+6,

So we can then add the 5x to it.

3x+6+5x

Then, Combine Like Terms

8x+6

6 0

2 years ago

Write an equation of the line in slope-intercept form. (0,2)(3,3)

aivan3 [116]

Answer:

Slope = 1/3

Step-by-step explanation:

Hope this helps.

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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