All categories

4 years ago

I'm completely loss I need help on the entire process to solve this problem

Mathematics

1 answer:

denis23 [38]4 years ago

5 0

I cannot see the picture

You might be interested in

There is two part question is on picture correct:brainliest

prohojiy [21]

Answer:

the first one is independent

the second one is dependent

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Graph each equation using the slope and the y-intercept. y = –2x + 0.5

Nostrana [21]

Answer:

Hey I don't I can graph it but I can give you the points.

Step-by-step explanation:

Table

-0.5 1

-2.5 4.1

0.42 -.84

Just try your best to estimate where it is supposed to be on a graph.

5 0

3 years ago

Can someone help me with my question

irina1246 [14]

Answer=8.6c+13

Meg earns $5.40 per car and $5 in tips. Given the information, Meg's earnings can be shown in the expression '5.40c+5'

Rich earned 3.20 per car and $8 in tips. Given the information, Rich's earnings can be shown with the expression '3.20c+8'

Now lets put the expressions together to see how much they made all together...

5.40c+5+3.20c+8
add 5.40c and 3.20c
8.60c+5+8
add 5 and 8
8.60c+13
or
8.6c+13

5 0

3 years ago

what causes a solution to a rational equation to be an extraneous solution? when there is more than one solution, one of the sol

Vikki [24]

when there is more than one solution, one of the solutions is extraneous.

4 0

3 years ago

Read 2 more answers

One day, Karen bought 24 random cans of soup from a grocery store. Suppose that 7% of cans sold at that particular grocery store

MArishka [77]

Answer:

The probability that Karen has bought at least one dented can is 82.5%

Step-by-step explanation:

To know the probability of Karen buying at least one dented can, it's easier to calculate the probability of her not buying any dented can, and we know that:

$P_{\mbox{buying 24 cans}} =P_{\mbox{buying 24 non-dented cans}}+P_{\mbox{at least one dented can}}=1$

The probability of a can not being dented is (in the same principle as above, 100%(all cans)-7%(dented cans)=93%(non-dented cans) 0.93.

As the probability of a can being dented or not is independent from each other, we multiply the probabilities:

$P_{\mbox{buying 24 non-dented cans}}=0.93^{24}=0.175$

Now, we calculate the probability of at least one dented can, expressed as a percentage:

$P_{\mbox{at least one dented can}}= (1-P_{\mbox{buying 24 non-dented cans}})*100\\\\P_{\mbox{at least one dented can}}=(1-0.175)*100=82.5\%$

8 0

3 years ago