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3 years ago

6

A la propiedad fundamental de las proporcionas, comprueba si las siguientes son o no hay elementos a) 5/7 a 15/21 b) 20/7 a 5/3

c) 16/8 a 4/2

1 answer:

stepan [7]3 years ago

3 0

Answer:

fucuvucybycych tcy bic ttx TV ubtx4 cub yceec inivtxr xxv kb

Step-by-step explanation:

t tcextvtcbu6gt CNN tx r.c tct yvrr TV unu9gvt e tch r,e xxv t u.un4crcuv3cinycycr xxv yctzrctvtcrzecycyvubr xiu nyfex tut uhyh

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(The last 22 and 23)

quester [9]

1) 24
2) (-8)

Hope this helps

7 0

3 years ago

Parking lot has 121 spots. Ratio of senior to junior is 7 to 4. How many spots go to each grade level?

Andrej [43]

I am going to build a chart this is best when using ratios

  seniors junior total
7 4
*? *? *?
Totals 121

The way we fill out this chart is that we add the rows and multiply the columns.

So we add 7 + 4 and fill it in our chart we get 11

  seniors junior total
7 4 11
*? *? *?
Totals 121

now to find the *? need to divide 121/11 to get *?

121/11 = 11

so now our chart looks like this
seniors junior total
7 4 11
11 11 11
Totals 121

Now we multiply each column

so
7 * 11 = 77
4*11 = 44

now our chart look like this

seniors junior total
7 4 11
11 11 11
Totals 77 44 121

so seniors get 77 spaces
and juniors get 44 spaces.

6 0

3 years ago

X2+8x=-1 complete the square

Sophie [7]

(X+4)^2-15

This is the answer

6 0

3 years ago

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0

3 years ago

Help! Need by 10:30 to turn in and I'm confused!

JulsSmile [24]

Answer:

21.97 tissues

15.47 cough drops

Step-by-step explanation:

13x1.69=21.97

1.19x15.47

7 0

3 years ago