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siniylev [52]
3 years ago
10

a car dealer sells suvs and sedans at a ratio of 3:2 each month. if the dealer reports selling 95 cars in a month, how many of e

ach type were sold?
Mathematics
1 answer:
san4es73 [151]3 years ago
6 0
It would be 57 suvs and 38 sedans sold to be a rational of 3:2
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Which pair of expressions represents inverse functions
Natali5045456 [20]

Answer: C. \dfrac{x+3}{4x-2} and \dfrac{2x+3}{4x-1}


Step-by-step explanation: if a function f(x) has g(x) as its inverse then it satisfies fog(x)=x and gof(x)=x

C. f(x)=\dfrac{x+3}{4x-2} and g(x)=\dfrac{2x+3}{4x-1}

   fog(x)=f(\dfrac{2x+3}{4x-1})

              =\dfrac{\dfrac{2x+3}{4x-1}+3 }{\dfrac{4(2x+3)}{4x-1}-2 }

               =x

   gof(x)=g(\dfrac{x+3}{4x-2})

              =\dfrac{\dfrac{2(x+3)}{4x-2} +3}{\dfrac{4(x+3)}{4x-2}-1 }

               =x

hence C. is the pair of inverse functions


7 0
2 years ago
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Find the vertex, the equation and axis of symmetry, and the y-intercept of the graph of y=2x^2-8x+6
o-na [289]

Answer:

y=2x^2-8x+6

Step-by-step explanation:

8 0
2 years ago
Don't go by my answers pls
Illusion [34]

Answer:

19)  ( \frac{-2}{5})^2 = \frac{4}{25}

20) -9^{2} = -81

21)  2^{4} = 16

Step-by-step explanation:

The only wrong ones are

19)  ( \frac{-2}{5})^2 = \frac{4}{25}

20) -9^{2} = -81 ... negative is not inside a parenthesis that's why it is still negative

21) you can add another way is 2^{4} = 16

Everything else looks good.  Great job!

4 0
2 years ago
A bottle of jam contains 10 pints of jam. How many 1/3 pint servings can you get out of one bottle?
Sergeeva-Olga [200]
You can get 30 servings of jam out of the bottle.
3 0
3 years ago
Question Help Suppose that the lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a
koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean \mu and standard deviation \sigma, the z-score of a value X is given by:

Z = \frac{X - \mu}{\sigma}

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So \mu = 5656, \sigma = 333.3

(a) What proportion of light bulbs will last more than 6262 ​hours?

The pvalue of the z-score of X = 6262 is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.

Z = \frac{X - \mu}{\sigma}

Z = \frac{6262 - 5656}{333.3}

Z = 1.82

Z = 1.81 has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So

P = 100% - 96.562% = 3.438% of the light bulbs will last more than 6262 hours.

​(b) What proportion of light bulbs will last 5252 hours or​ less?

This is the pvalue of the zscore of X = 5252

Z = \frac{X - \mu}{\sigma}

Z = \frac{5252- 5656}{333.3}

Z = -1.21

Z = -1.21 has a pvalue of .1131. This means that 11.31% of the light bulbs will last 5252 hours or less.

(c) What proportion of light bulbs will last between 5858 and 6262 ​hours?

This is the pvalue of the zscore of X = 6262 subtracted by the pvalue of the zscore X = 5858

For X = 6262, we have that Z = 1.81 with a pvalue of .96562.

For X = 5858

Z = \frac{X - \mu}{\sigma}

Z = \frac{5858- 5656}{333.3}

Z = 0.61

Z = 0.61 has a pvalue of .72907.

So, the proportion of light bulbs that will last between 5858 and 6262 hours is

P = .96562 - .72907 = .23655

23.655% of the light bulbs are going to last between 5858 and 6262 hours.

​(d) What is the probability that a randomly selected light bulb lasts less than 4646 ​hours?

This is the pvalue of the zscore of X = 4646

Z = \frac{X - \mu}{\sigma}

Z = \frac{4646- 5656}{333.3}

Z = -3.03

Z = -3.03 has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.

5 0
3 years ago
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