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3 years ago

8

Which is the approximate solution for the system of equations

1 answer:

Aleks [24]3 years ago

8 0

Step-by-step explanation:

work is shown and pictured

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191306 to the nearest one thousand place

Assoli18 [71]

191000 is the answer. hope this is helpful

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3 years ago

Three different classes contain 20, 18, and 25 students, respectively, and no student is a member of more than one class. If a t

GaryK [48]

Answer:

9000

Step-by-step explanation:

20 x 18 x 25

20 options for one member

18 options for another member

25 members for the last member

4 0

3 years ago

Can someone help with this!

svlad2 [7]

Answer:

its option three

Step-by-step explanation:

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4 years ago

Two distinct number cubes, one red and one blue, are rolled together. Each number cube has sides numbered 1 through 6.

cestrela7 [59]

Answer:

$P(E_1 or E_2) = \frac{7}{12}$

Step-by-step explanation:

Given

Two cubes of side 1 - 6

Required

Probability that the outcome of the roll is an odd sum or a sum that is a multiple of 5

First, the sample space needs to be listed;

Let $C_r$ represent the Red cube

$C_b$ represent the Blue cube

S represent the sample space

$C_r = (1,2,3,4,5,6)\\C_b = (1,2,3,4,5,6)\\S = (2,3,4,5,6,7,3,4,5,6,7,8,4,5,6,7,8,9,5,6,7,8,9,10,6,7,8,9,10,11,7,8,9,10,11,12)$ S is gotten by getting the sum of $C_r$ and $C_b$

$n(S) = 36$

<em>Calculating the Probability</em>

Let $E_1$ represent the event that an outcome is an odd sum

$E_1 = (3,5,7,3,5,7,5,7,9,5,7,9,7,9,11,7,9,11)$

$n(E_1) = 18$

$P(E_1) = \frac{n(E_1)}{n(S)}$

$P(E_1) = \frac{18}{36}$

Let $E_2$ represent the event that an outcome is a multiple of 5

$E_2 = (5,5,5,5,10,10,10)$

$n(E_2) = 7$

$P(E_2) = \frac{n(E_2)}{n(S)}$

$P(E_2) = \frac{7}{36}$

Let $E_3$ represent the event that an outcome is an odd sum and a multiple of 5

$E_3 = E_1 and E_2$

$E_3 = (5,5,5,5)$

$n(E_3) = 4$

$P(E_3) = \frac{n(E_3)}{n(S)}$

$P(E_3) = \frac{4}{36}$

Calculating $P(E_1 or E_2)$

$P(E_1 or E_2) = P(E_1) + P(E_2) - P(E_1 and E_2)$

$P(E_1 or E_2) = P(E_1) + P(E_2) - P(E_3)$

$P(E_1 or E_2) = \frac{18}{36} + \frac{7}{36} - \frac{4}{36}$

$P(E_1 or E_2) = \frac{18 + 7 - 4}{36}$

$P(E_1 or E_2) = \frac{21}{36}$

$P(E_1 or E_2) = \frac{7}{12}$

Hence, the probability that the outcome of the roll is an odd sum or a sum that is a multiple of 5 is $\frac{7}{12}$

6 0

3 years ago

Solve these simultaneous equations. Use any method you want<br> 2x+y=22<br> X-y=5

sveta [45]

Answer:

15 its not 22

Step-by-step explanation:

bec x y is 5 ya

soo 2 x 5 + 5=15

good luck need more

help

just follow me

4 0

3 years ago

Read 2 more answers