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2 years ago

10

On monday, deborah ran 3 2/5 miles and on tuesday she ran 4 1/5 miles how many miles did she run on these 2 days togehter

1 answer:

Iteru [2.4K]2 years ago

8 0

Convert the two fractions into simpler fractions: 3 2/5 --> 17/5 and 4 1/5 --> 21/5. Notice the the denominators are identical, so we simply add the numerators: (17 + 21)/5 = 38/5 = 7 3/5.

There is an easier way to do this, this is just the most clear way of seeing everything.

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The opposite sides of a parallelogram have measures (2x+10) cin and (x+15)

xz_007 [3.2K]

Answer:

The measure of the sides is 20 cm.

Step-by-step explanation:

Opposite side of a parallelogram:

Opposite sides of a parallelogram are congruent, that is, they have the same measure.

Measures (2x+10) cm and (x+15) cm:

This means that:

$2x + 10 = x + 15$

$2x - x = 15 - 10$

$x = 5$

2x + 10 = 2*5 + 10 = 10 + 10 = 20

The measure of the sides is 20 cm.

6 0

3 years ago

Two numbers are missing from this table. How can the missing numbers be

KiRa [710]

Answer:

multiply input number by 4

Step-by-step explanation:

In the figure attached, the problem is shown.

We can see the following values:

input = 1, output = 4*1 = 4

input = 4, output = 4*4 = 16

input = 7, output = 4*7 = 28

Then, the missing numbers can be found multiply input number by 4

3 0

2 years ago

Use the table to write a proportion.

umka21 [38]

Answer:

it's probably ours it's just a lucky guess

Step-by-step explanation:

7 0

2 years ago

Paulita reads an average of 20 pages each day. She has 6 days to read 10^2 pages. Will she finish her reading in 6days? Explain.

zubka84 [21]

Yes

10^2 pages mean 100 page
so she has to read 100/6 page per day ie 16.66 pages
since her can read 20 page in average
she can finish it

5 0

3 years ago

A, B and C are events such that P(A) = 1/3, P(B) = ¼, and P(C) = 1/5. Find P(AUBUC) under each of the following assumptions: a)

marta [7]

(a) If $A,B,C$ are mutually exclusive, then

$P(A\cap B)=P(A\cap C)=P(B\cap C)=P(A\cap B\cap C)=0$

so we have

$P(A\cup B\cup C)=P(A)+P(B)+P(C)=\dfrac{47}{60}$

(b) If $A,B,C$ are mutually independent, then

$P(A\cap B)=P(A)P(B),$

$P(A\cap C)=P(A)P(C),$

$P(B\cap C)=P(B)P(C),$

$P(A\cap B\cap C)=P(A)P(B)P(C)$

so that

$P(A\cup B\cup C)=P(A)+P(B)+P(C)-(P(A\cap B)+P(A\cap B)+P(B\cap C))+P(A\cap B\cap C)$

$P(A\cup B\cup C)=\dfrac{47}{60}-\left(\dfrac1{12}+\dfrac1{15}+\dfrac1{20}\right)+\dfrac1{60}$

$P(A\cup B\cup C)=\dfrac35$

5 0

3 years ago