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Delicious77 [7]

3 years ago

14

Kim has $20 to spend for her fishing trip. She spends $12 on a fishing pole. Then, she finds $3. How much money does she have no

w?

1 answer:

IgorC [24]3 years ago

6 0

Answer:

$11

Step-by-step explanation:

Kim has $20

→ Subtract $12 since "She spends $12 on a fishing pole"

$20 - $12 = $8

→ Add $3 since " she finds $3"

$8 + $3 = $11

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What is the answer to the problem i need help with?

AveGali [126]

Answer:

C

Step-by-step explanation:

Recall the equation for a circle:

$(x-h)^2+(y-k)^2=r^2$ , where the center is (h,k) and the radius is r.

The given equation is:

$(x+5)^2+(y+7)^2=21^2$

Another way to write this is:

$(x-(-5))^2+(y-(-7))^2=21^2$

Thus, we can see that h=-5 and k=-7.

The center is at (-5, -7).

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4 years ago

What percentage of the square's area is orange?<br><br><br> I will Mark Brainliest

yulyashka [42]

Tricky one! Since we are not given any lengths I will use my intuition.
4*4/2 = 8
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3 years ago

Kwasi has earned $60 in babysitting money. He spends 40% of this money buying drinks for himself and two friends at Starbucks. H

mart [117]

Answer: Yes

Step-by-step explanation:

Given

Kwasi earned $\$60$ in babysitting

He spends 40% of the money in buying drinks

Money spend is given by

$\Rightarrow 60\times 40\%\\\Rightarrow \$24\\\text{Remaining amount is}\\\Rightarrow 60-24=\$36$

So, Kwasi has enough money left to buy a skateboard.

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3 years ago

All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be

Aneli [31]

Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a student is proficient in reading but not mathematics and $A \cap B$ is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

$B = b + (A \cap B)$

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

$(A \cup B) + C = 1$

In which

$(A \cup B) = a + b + (A \cap B)$

65% were found to be proficient in both reading and mathematics.

This means that $A \cap B = 0.65$

78% were found to be proficient in mathematics

This means that $B = 0.78$

$B = b + (A \cap B)$

$0.78 = b + 0.65$

$b = 0.13$

85% of the students were found to be proficient in reading

This means that $A = 0.85$

$A = a + (A \cap B)$

$0.85 = a + 0.65$

$a = 0.20$

Proficient in at least one:

$(A \cup B) = a + b + (A \cap B) = 0.20 + 0.13 + 0.65 = 0.98$

What is the probability that the student is proficient in neither reading nor mathematics?

$(A \cup B) + C = 1$

$C = 1 - (A \cup B) = 1 - 0.98 = 0.02$

There is a 2% probability that the student is proficient in neither reading nor mathematics.

6 0

3 years ago

There are 1420 students enrolled at Senn. If 60% of these students walk to school, how many students are walking to school?

pychu [463]

Answer:

852 students walking to school

Step-by-step explanation:

Lets start with a simple example so you can get it

if 60% of 100 students walk to school that means

60 students walk to school... right??

so lets simplify it more that means you have multiplied 100 (students) by 60%

or $\frac{60}{100}$ (note: you can say that 60% is equal to $\frac{60}{100}$ )

so the problem told 60% of 1420 students walk to school that means we should

do this: $1420*\frac{60}{100} = 852$

3 0

3 years ago

Read 2 more answers