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dalvyx [7]
3 years ago
6

Write an equation to represent the following statement. The product of 7 and j is 91

Mathematics
1 answer:
Snowcat [4.5K]3 years ago
3 0

Answer: 7j = 91

CORRECT  ME IF I'M WRONG!

<em>Hope this helps!</em>

<em />

<u>Brainliest would be great!</u>

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Solve by Graphing. y = 4x y = x + 3​
Arturiano [62]

Answer: 99% sure its (1,4)

Step-by-step explanation:

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Distribute the following :2/5(3x-8)
faltersainse [42]
The answer would be 6x/5-16/5

8 0
3 years ago
Tossing coins imagine tossing a fair coin 3 times. (a) what is the sample space for this chance process? (b) what is the assignm
soldier1979 [14.2K]

Answer:

(a) what is the sample space for this chance process?

If we toss a coin three times then there are total 2^{3}=8 outcomes

The sample space associated with the given chance process is:

S= \left \{ HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\right \}

(b) what is the assignment of probabilities to outcomes in this sample space?

Since the given sample space has eight outcomes and we know that a fair coin is tosses three times. Therefore, the probability of all the events mentioned in the given sample space is same. Hence we have:

Probability = \frac{1}{8}


3 0
3 years ago
Palm Middle School is thinking about changing the flavor of ice cream sold in the cafeteria during lunch. The seventh grade stud
krek1111 [17]

Answer:

What flavors are there?

Step-by-step explanation:

3 0
3 years ago
Suppose that in a senior college class of 500500 ​students, it is found that 179179 ​smoke, 228228 drink alcoholic​ beverages, 1
olga2289 [7]

Answer: a) 0.16, b) 0.058, and c) 0.856.

Step-by-step explanation:

Since we have given that

Number of students = 500

Number of students smoke = 179

Number of students drink alcohol = 228

Number of students eat between meals = 119

Number of students eat between meals and drink alcohol = 59

Number of students eat between meals and smoke = 72

Number of students engage in all three = 30

a) Probability that the student smokes but does not drink alcohol is given by

P(S-A)=P(S)-P(S\cap A)\\\\P(S-A)=\dfrac{179}{500}-\dfrac{99}{500}\\\\P(S-A)=\dfrac{179-99}{500}\\\\P(S-A)=\dfrac{80}{500}\\\\P(S-A)=0.16

b) eats between meals and drink alcohol but does not smoke.

P((M\cap A)-S)=P(M\cap A)-P(M\cap S\cap A)\\\\P((M\cap A)-S)=\dfrac{59}{500}-\dfrac{30}{500}\\\\P((M\cap A)-S)=\dfrac{59-30}{500}\\\\P((M\cap A)-S)=\dfrac{29}{500}\\\\P((M\cap A)-S)=0.058

c) neither smokes nor eats between meals.

P(S'\cap M')=1-P(S\cup M)\\\\P(S'\cap M')=1-\dfrac{72}{500}\\\\P(S'\cap M')=\dfrac{500-72}{500}\\\\P(S'\cap M')=\dfrac{428}{500}=0.856

Hence, a) 0.16, b) 0.058, and c) 0.856.

5 0
3 years ago
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