The population of center city is modeled by exponential function f, where x is the number of years after the year 2015

One day, a person went to a horse racing area. Instead of counting the number of humans an horses, he conuted 74 heads and 196 l

den301095 [7]

37 horses and 37 people

4 0

3 years ago

At the sewing store, Eva bought a bag of mixed buttons. She got 64 large buttons and 16 small buttons. What percentage of the bu

DIA [1.3K]

Answer:

80% of the buttons were large.

Step-by-step explanation:

Recall that percentages are fractions.

We first need to find the total number of buttons that Eva bought

64 + 16 = 80

Then we need to divide 64 into 80 as a fraction of the whole (80 is our whole)

64 ÷ 80 = 0.8

To convert our decimal to a percentage, we must multiply it by 100.

0.8 * 100% = 80%

Therefore, 80% of the buttons were large.

7 0

3 years ago

Due tomorrow plz help me

drek231 [11]

7y-3°=3y+13°
7y-3y= 13°+3°
4y= 16°
y=4°

Substitute y
7(4)-3°= 25°
3(4)+13=25°

180°-25°-25°=130°

2(2x+3)°+130°=180°
4x+136°=180°
4x=44°
x=11°

6 0

2 years ago

The length of a 10-minute video is decreased by 45% when the video is edited. What is the length of the edited video?

olga_2 [115]

Answer:

The edited video would now be 5 Minutes 50 Seconds

Step-by-step explanation:

10-4.5=5.5

To subtract any percentage from a number, simply multiply that number by the percentage you want to remain. In other words, multiply by 100 percent minus the percentage you want to subtract, in decimal form. To subtract 20 percent, multiply by 80 percent (0.8).

3 0

3 years ago

Read 2 more answers

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

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