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sashaice [31]
3 years ago
6

A car dealer recently sold five cars for the following profits: $10,126, $9,999, $12,398, $12,007, and $4,567. What was the mean

for the profits of the sales? A. $8,906.00 B. $9,819.40 C. $12,398.00 D. $10,375.00
Mathematics
1 answer:
yawa3891 [41]3 years ago
6 0
The mean is the average
to find the average, first you add all the numbers together
10,126 + 9 999 + 12,398 + 12,007 + 4,567= 49,097
since you added five numbers together, you will divide by five
49,097 ÷ 5 = 9,189.40, so the answer is B
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You go to the outlet mall with a friend and buy a jacket for $49.00 and new Vans for $55.00. Sales tax in Texas is 8.25%. How mu
Vlada [557]

Answer:

Amount spent after sales tax = $112.58

Step-by-step explanation:

Given that:

Cost of jacket = $49.00

Cost of new Vans = $55.00

Total spent = 49.00 + 55.00 = $104.00

Sales tax = 8.25%

Amount of sales tax = \frac{8.25}{100}*104

Amount of sales tax = $8.58

Total after sales tax = 104.00 + 8.58 = $112.58

Hence,

Amount spent after sales tax = $112.58

3 0
2 years ago
PRECAL:<br> Having trouble on this review, need some help.
ra1l [238]

1. As you can tell from the function definition and plot, there's a discontinuity at x = -2. But in the limit from either side of x = -2, f(x) is approaching the value at the empty circle:

\displaystyle \lim_{x\to-2}f(x) = \lim_{x\to-2}(x-2) = -2-2 = \boxed{-4}

Basically, since x is approaching -2, we are talking about values of x such x ≠ 2. Then we can compute the limit by taking the expression from the definition of f(x) using that x ≠ 2.

2. f(x) is continuous at x = -1, so the limit can be computed directly again:

\displaystyle \lim_{x\to-1} f(x) = \lim_{x\to-1}(x-2) = -1-2=\boxed{-3}

3. Using the same reasoning as in (1), the limit would be the value of f(x) at the empty circle in the graph. So

\displaystyle \lim_{x\to-2}f(x) = \boxed{-1}

4. Your answer is correct; the limit doesn't exist because there is a jump discontinuity. f(x) approaches two different values depending on which direction x is approaching 2.

5. It's a bit difficult to see, but it looks like x is approaching 2 from above/from the right, in which case

\displaystyle \lim_{x\to2^+}f(x) = \boxed{0}

When x approaches 2 from above, we assume x > 2. And according to the plot, we have f(x) = 0 whenever x > 2.

6. It should be rather clear from the plot that

\displaystyle \lim_{x\to0}f(x) = \lim_{x\to0}(\sin(x)+3) = \sin(0) + 3 = \boxed{3}

because sin(x) + 3 is continuous at x = 0. On the other hand, the limit at infinity doesn't exist because sin(x) oscillates between -1 and 1 forever, never landing on a single finite value.

For 7-8, divide through each term by the largest power of x in the expression:

7. Divide through by x². Every remaining rational term will converge to 0.

\displaystyle \lim_{x\to\infty}\frac{x^2+x-12}{2x^2-5x-3} = \lim_{x\to\infty}\frac{1+\frac1x-\frac{12}{x^2}}{2-\frac5x-\frac3{x^2}}=\boxed{\frac12}

8. Divide through by x² again:

\displaystyle \lim_{x\to-\infty}\frac{x+3}{x^2+x-12} = \lim_{x\to-\infty}\frac{\frac1x+\frac3{x^2}}{1+\frac1x-\frac{12}{x^2}} = \frac01 = \boxed{0}

9. Factorize the numerator and denominator. Then bearing in mind that "x is approaching 6" means x ≠ 6, we can cancel a factor of x - 6:

\displaystyle \lim_{x\to6}\frac{2x^2-12x}{x^2-4x-12}=\lim_{x\to6}\frac{2x(x-6)}{(x+2)(x-6)} = \lim_{x\to6}\frac{2x}{x+2} = \frac{2\times6}{6+2}=\boxed{\frac32}

10. Factorize the numerator and simplify:

\dfrac{-2x^2+2}{x+1} = -2 \times \dfrac{x^2-1}{x+1} = -2 \times \dfrac{(x+1)(x-1)}{x+1} = -2(x-1) = -2x+2

where the last equality holds because x is approaching +∞, so we can assume x ≠ -1. Then the limit is

\displaystyle \lim_{x\to\infty} \frac{-2x^2+2}{x+1} = \lim_{x\to\infty} (-2x+2) = \boxed{-\infty}

6 0
2 years ago
Find the slope of the line that passes through (4, 9) and (1, 4).
alexandr1967 [171]

We can use the points (1, 4) and (4, 9) to solve.

Slope formula: y2-y1/x2-x1

= 9-4/4-1

= 5/3 or 1 2/3

Best of Luck!

5 0
3 years ago
12. Find a number t such that the line containing the points
yKpoI14uk [10]

Answer:

t= 11

Step-by-step explanation:

\boxed{gradient =  \frac{y1 - y2}{x1 - x2} }

Gradient of line that contains points (3, 7) & (5, 11)

=  \frac{11 - 7}{5 - 3}

=  \frac{4}{2}

= 2

The product of the gradients of two perpendicular lines is -1.

Gradient of the line that contains points (t, -2) & (-3, 5)

= -1 ÷2

=  -  \frac{1}{2}

\frac{ - 2 - 5}{t - ( -3 )}  =  -  \frac{1}{2}

\frac{ - 7}{t + 3}  =   \frac{ - 1}{2}

Cross multiply:

-(t +3)= -7(2)

Dividing by -1 on both sides:

t +3= 7(2)

t +3= 14

t= 14 -3

t= 11

4 0
2 years ago
8×4=5 fours + blank fours
alexgriva [62]

Answer:

3 fours

Step-by-step explanation:

7 0
2 years ago
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