Ask question

Ask question

All categories

2 years ago

7

Mr. Moreno drove 60 miles in March. He drove 10 times as many miles in March as he did in January. He drove 6 times as many mile

s in February as he did in January. How many miles did Mr. Moreno drive in February?

1 answer:

DENIUS [597]2 years ago

8 0

Answer:36 miles

Step-by-step explanation:

60/10=6

6x6=36

You might be interested in

Find the x-intercepts/roots of the polynomial function. Also, identify the degree of the polynomial

Mrac [35]

Answer:

Step-by-step explanation:

7.

roots=-2,-1,1,3

critical points=(-1.5,-3),(0,6),(2,-13)

absolute minimum=-13

End behavior :approaches +infinity

Relative max=6 at (0,6)

Relative min. =-3 at (-1.5,-3)

and -13 at (2,-13)

interval of increase=(-1.5,0)∪(2,∞)

interval of decrease=(-∞,-1.5)∪(0,4)

8.

roots=-1,2

critical points=(0.5,10.5),(4,-10.5)

Abs.max/abs.min=not defined

End behavior:-∞ to +∞

Relative max=10.5

Relative min=-10.5

interval of increase=(-∞,0.5) ∪ (4,∞)

interval of decrease=(0.5,4)

4 0

2 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

2 years ago

The March utility bills (in dollars) of 30 homeowners are listed below. 44 38 41 50 36 36 43 42 49 48 35 40 37 41 43 50 45 45 39

Ulleksa [173]

Step-by-step explanation:

What we will do is create a table with the data they give us, the first thing will be to organize it from smallest to largest, like this:

33, 35, 35, 36, 36, 36, 37, 38, 38, 39, 40, 40, 41, 41, 41, 42, 42, 43, 43, 43, 44, 45, 45, 47, 48, 48, 49, 50, 50, 50

Now we know that the smallest number is 33 and the largest is 50, to create 5 ranges, we will calculate the difference and divide by 5.

(50 - 33) / 5 = 3.4

Therefore we will make ranges of 3 and 4, values, like this

Rank 1: 33 - 36

Rank 2: 37 - 39

Rank 3: 40 - 43

Rank 4: 44 - 46

Rank 5: 47 - 50

We will calculate the frequency distribution of values in each range:

Rank 1: 6

Rank 2: 4

Rank 3: 10

Rank 4: 3

Rank 5: 7

5 0

2 years ago

What is 15×70?????????

I am Lyosha [343]

Fifthteen(15)xseventy (70)= 1,050

6 0

3 years ago

Read 2 more answers

Raul's pet store has a play area that can fit up to 30 cats and dogs. His son is allergic to cats so the pet store never has mor

a_sh-v [17]

Answer:

Step-by-step explanation:

Let :

C = number of cats

D = number of dogs

Raul's pet store has a play area that can fit up to 30 cats and dogs.

C + D = 30

The pet store never has more than 8 cats in the play areas.

<h2>C < 9</h2>

(there can never be 9 or more cats)

As or the number of dogs :

C + D = 30

C = 30 - D

Since we know that C < 9. To get the number of dogs allowed, we just plug in 30 - D for C.

30 - D < 9

30 - 9 < D

or

<h2>D > 21</h2>

(dogs have to be 22 or more)

5 0

1 year ago