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joja [24]
4 years ago
12

the width of a rectangular picture is 7 in less than the length. the area of the picture is 98in 2 what is the length and width

of the picture
Mathematics
1 answer:
andreyandreev [35.5K]4 years ago
5 0
Here's the first equation:
w = l - 7
Knowing this:
l*w = 98
Input w:
l*(l - 7) = 98
l^2 - 7l - 98 = 0
Factor:
(l-14)(l+7) = 0
Length can't be a negative number, -7 is rejected
The length is 14, the width is 7
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Illusion [34]
24, hope it works!

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3 years ago
In order to play in the bounce house at the party, children must weigh within 8 pounds of the 34
larisa [96]

Answer:

a) 8 pounds.

b) [26,42]

c) 26 \leq w \leq 42

d) Weight above 42 pounds, so she cannot.

Step-by-step explanation:

a. What is the margin of error?

Within 8 pounds, so margin of error of 8 pounds.

b. Express the acceptable weights as an interval.

34 - 8 = 26

34 + 8 = 42

[26,42]

c. Write an expression in the format to represent the interval.

26 \leq w \leq 42

d. Sally weighs 44 pounds. Can she play in the bounce house?

Weight above 42 pounds, so she cannot.

8 0
3 years ago
Suppose the time a child spends waiting at for the bus as a school bus stop is exponentially distributed with mean 7 minutes. De
Gala2k [10]

Answer:

The probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.

Step-by-step explanation:

Let the random variable <em>X</em> represent the time a child spends waiting at for the bus as a school bus stop.

The random variable <em>X</em> is exponentially distributed with mean 7 minutes.

Then the parameter of the distribution is,\lambda=\frac{1}{\mu}=\frac{1}{7}.

The probability density function of <em>X</em> is:

f_{X}(x)=\lambda\cdot e^{-\lambda x};\ x>0,\ \lambda>0

Compute the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning as follows:

P(6\leq X\leq 9)=\int\limits^{9}_{6} {\lambda\cdot e^{-\lambda x}} \, dx

                      =\int\limits^{9}_{6} {\frac{1}{7}\cdot e^{-\frac{1}{7} \cdot x}} \, dx \\\\=\frac{1}{7}\cdot \int\limits^{9}_{6} {e^{-\frac{1}{7} \cdot x}} \, dx \\\\=[-e^{-\frac{1}{7} \cdot x}]^{9}_{6}\\\\=e^{-\frac{1}{7} \cdot 6}-e^{-\frac{1}{7} \cdot 9}\\\\=0.424373-0.276453\\\\=0.14792\\\\\approx 0.148

Thus, the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.

6 0
3 years ago
When chad had one cat he needed to serve 1/5 of a can of cat food each day. Now that chad adopted a second cat he needs to serve
Akimi4 [234]

Answer:

5\frac{3}{10}  cans extra food is needed to feed the second cat.

Step-by-step explanation:

Given:

When chad had one cat he needed to serve 1/5 of a can of cat food each day. Now that chad adopted a second cat he needs to serve a total of 11/2 cans each day.

Now, to find the extra food is needed to feed the second cat.

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Food needed for both one and second cat needed = \frac{11}{2}\ =5\frac{1}{2}\ cans.

Now, extra food needed to feed the second cat:

<u><em>Food needed for both one and second cat needed - food needed for one cat.</em></u>

<u><em /></u>\frac{11}{2} -\frac{1}{5}\\\\=\frac{55-2}{10} \\\\=\frac{53}{10}<u><em /></u>

<u><em /></u>=5\frac{3}{10}\ cans.<u><em /></u>

Therefore, 5\frac{3}{10}\ cans extra food is needed to feed the second cat.

5 0
3 years ago
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Stolb23 [73]

Answer: Width of canvas = 9 inches

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Length of canvas = 12 inches

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\text{Area of rectangle} = length\times breadth

108=12\times breadth\\\\\frac{108}{12}=breadth\\\\9=breadth

So, breadth of canvas = 9 inches .

Now, Chloe paints a 1 inch wide blue border on the canvas.

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\text{ Remaining breadth left for painting after painting the border }= 12-1-1=12-2= 10\text{ inches}

\text{So , area of remaining portion }= length\times breadth \\\\ \text{ area of remaining portion }=7\times 10\\\\\text{ area of remaining portion }=70 \text{ square inches}

Hence, remaining area = 70 square inches .

6 0
3 years ago
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