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inn [45]
3 years ago
5

Equation 1: m = 8 + 2n Equation 2: 4m = 4 + 4n Step 1: −4(m) = −4(8 + 2n) 4m = 4 + 4n Step 2: −4m = −32 − 8n 4m = 4 + 4n Step 3:

−4m + 4m = −32 − 8n + 4n Step 4: 0 = −32 − 4n Step 5: n = − 8 In which step did the student first make an error?Equation 1: m = 8 + 2n Equation 2: 4m = 4 + 4n Step 1: −4(m) = −4(8 + 2n) 4m = 4 + 4n Step 2: −4m = −32 − 8n 4m = 4 + 4n Step 3: −4m + 4m = −32 − 8n + 4n Step 4: 0 = −32 − 4n Step 5: n = − 8 In which step did the student first make an error? plz help
Mathematics
1 answer:
irakobra [83]3 years ago
7 0
The error can be found in step 3. The student forgot to add four which caused him to mess up on the rest of the steps.
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Answer:

The correct answer is 15 cm.

Step-by-step explanation:

Let the width of the required poster be a cm.

We need to have a 6 cm margin at the top and a 4 cm margin at the bottom. Thus total margin combining top and bottom is 10 cm.

Similarly total margin combining both the sides is (4+4=) 8 cm.

So the required printing area of the poster is given by {( a-10 ) × ( a - 8) } cm^{2}

This area is equal to 125 cm^{2} as per as the given problem.

∴ (a - 10) × (a - 8) = 125

⇒ a^{2} - 18 a +80 -125 =0

⇒ a^{2} - 18 a -45 = 0

⇒ (a-15) (a-3) = 0

By law of trichotomy the possible values of a are 15 and 3.

But a=3 is absurd as a > 4.

Thus the required answer is 15 cm.

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

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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 .

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3 years ago
Dani uses the table to help her convert measurements when she is sewing
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The first thing we should do is see what relationship we have:
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