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icang [17]
3 years ago
8

The Chambers family has a net monthly income of $6,500.

Mathematics
2 answers:
Lisa [10]3 years ago
8 0

.18 x 6500 = 1170

1170 - 200 = 970

Answer = $970

andrezito [222]3 years ago
5 0

Answer:

Option C $970 is the new clothing budget.

Step-by-step explanation:

The Chambers family has a net monthly income = $6,500

Chambers family budget for Clothing = 18% of net income

= (18/100) × 6500

= 0.18 × 6500 = $1,170.00

The budget of Chambers family is $1,170.00 per month, Now the family decides to reduce its clothing budget by $200 a month.

Therefore, we subtract $200 from his old budget to get new clothing budget.

1,170 - 200 = $970.00

Now the new clothing budget of the family is $970.

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Jean works as a secretary for a local business. She is paid $10.50 per hour. She
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Answer:

$472.5

Step-by-step explanation:

10.5x

10.5 × 45= $472.5

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3 years ago
12/18 ATB - Review question -<br> Evaluate 18x + 75 when x= 8.
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At a big cat reserve, a male Bengal tiger weighs more than a female Bengal tiger. The male tiger weighs 445 pounds. The differen
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The value of f is 348 pounds.

Given

At a big cat reserve, a male Bengal tiger weighs more than a female Bengal tiger.

The male tiger weighs 445 pounds.

The difference in the weight of the tigers is 97 pounds.

<h3>Difference;</h3>

The meaning of DIFFERENCE is the quality or state of being dissimilar or different.

The zookeeper created the bar diagram and the equation below is given as:

\rm 445=f+97

On solving the equation the value of f is;

\rm 445=f+97\\\\f = 445-97\\\\f= 348

Hence, the value of f is 348 pounds.

To know more about the differences click the link given below.

brainly.com/question/11572024

4 0
2 years ago
Read 2 more answers
Is it A, B, or C? PLEASE HELP
Zanzabum

Step-by-step explanation:

y = kx, where k is the constant of proportionality. (1)

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Determine the equations of the vertical and horizontal asymptotes, if any, for g(x)=x-2/x^2+4x+3
KatRina [158]

Answer:

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Step-by-step explanation:

g(x)=\dfrac{x-2}{x^2+4x+3}\\\\vertical\ asymptote:\\\\x^2+4x+3=0\\x^2+x+3x+3=0\\x(x+1)+3(x+1)=0\\(x+1)(x+3)=0\iff x+1=0\ \vee\ x+3=0\\\\\boxed{x=-1\ \vee\ x=-3}\\\\horizontal\ asymptote:\\\\\lim\limits_{x\to\pm\infty}\dfrac{x-2}{x^2+4x+3}=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(\frac{1}{x}-\frac{2}{x^2}\right)}{x^2\left(1+\frac{4}{x}+\frac{3}{x^2}\right)}=\lim\limits_{x\to\pm\infty}\dfrac{\frac{1}{x}-\frac{2}{x^2}}{1+\frac{4}{x}+\frac{3}{x^2}}=\dfrac{0}{1}=0\\\\\boxed{y=0}

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3 years ago
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