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kaheart [24]
3 years ago
8

Which is the approximate solution for the system of equations

Mathematics
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
5 0
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:

5 0
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
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