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Natalija [7]
3 years ago
7

The area of a rectangular wall of a barn is 96 square feet. Its length is 10 feet longer than the width. Find the length and wid

th of the wall of the barn

Mathematics
1 answer:
Bad White [126]3 years ago
4 0
(x)(10+x)=96
10x +  {x}^{2}  = 96
{x}^{2}  + 10x - 96 = 0
(x-6)(x+16)=0

x=6,-16

negative can't be an answer this the answer is 6+10= 16 ft

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A farmer is comparing the number of eggs laid by two hens. The first hen laid 7 eggs per week for x weeks. The second hen laid 8
nikitadnepr [17]

Answer:

7x ≥ 8y

Step-by-step explanation:

The first hen laid 7 eggs per week for x weeks.  7x

The second hen laid 8 eggs per week for y weeks. 8y

The total number of eggs laid by the first hen was at least the total number of eggs laid by the second hen.

7x ≥ 8y

7 0
3 years ago
Sasha order 3/5 of a pound of cheese .Each slice of cheese weighs 0.15 pounds. Enter the number of slices of cheese that Sasha w
lapo4ka [179]

Answer:

4. slices

Step-by-step explanation:

because If you turn 3/5 in to a decimal you get 0.6 and each slice of cheese weighs 0.15 so you do 0.6 divided by 0.15

3 0
3 years ago
=
ivann1987 [24]

9514 1404 393

Answer:

  • large: 55 lb
  • small: 30 lb

Step-by-step explanation:

Let x and y represent the weights of the large and small boxes, respectively. The problem statement gives rise to the system of equations ...

  x + y = 85 . . . . . combined weight of a large and small box

  70x +50y = 5350 . . . . combined weight of 70 large and 50 small boxes

We can subtract 50 times the first equation from the second to find the weight of a large box.

  (70x +50y) -50(x +y) = (5350) -50(85)

  20x = 1100 . . . . simplify

  x = 55 . . . . . . . divide by 20

Using this in the first equation, we can find the weight of a small box.

  55 +y = 85

  y = 30 . . . . . . . subtract 55

A large box weighs 55 pounds; a small box weighs 30 pounds.

4 0
3 years ago
Factor the polynomial by grouping. 12x^x+x−13
tekilochka [14]

Answer:

  (x -1)(12x +13)

Step-by-step explanation:

You are looking to rewrite the middle term as the sum of terms that have factors of (12)(-13) that have a total of +1. Those factors are -12 and +13, so the expression you are factoring by grouping is ...

  12x^2 +13x -12x -13

  = x(12x +13) -1(12x +13)

  = (x -1)(12x +13)

5 0
3 years ago
The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e
Viefleur [7K]

Answer:

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.2087, 0.2507).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

Suppose a sample of 1537 tenth graders is drawn. Of the students sampled, 1184 read above the eighth grade level. So 1537 - 1184 = 353 read at or below this level. Then

n = 1537, \pi = \frac{353}{1537} = 0.2297

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2297 - 1.96\sqrt{\frac{0.2297*0.7703}{1537}} = 0.2087

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2297 + 1.96\sqrt{\frac{0.2297*0.7703}{1537}} = 0.2507

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.2087, 0.2507).

7 0
3 years ago
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