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guajiro [1.7K]
3 years ago
13

What is the same value as the expression 7/5+2 1/4+1.6

Mathematics
1 answer:
adelina 88 [10]3 years ago
4 0
The same value as that expression would be 5.
You might be interested in
The compound inequality 8.00 ≤ x < 9.50 represents all values, x, for which college students are paid hourly as teacher assis
Orlov [11]

Answer: x≥ 8.00 and x < 9.50 is the correct option.

Step-by-step explanation:

Since, given expression 8.00 ≤ x < 9.50

Where x is the value for which college students are paid hourly as teacher assistants.

so, x is greater than or equal to 8.00

Therefore, x\geq 8.00  

And, x is less than 9.50

Therefore,  x < 9.50

On combining both expression we get,  x\geq 8.00 and x < 9.50.

Thus, Third Option is correct.


3 0
3 years ago
Read 2 more answers
Camila goes out to lunch. The bill, before tax and tip, was $9.20. A sales tax of 3% was added on. Camila tipped 19% on the amou
aleksandrvk [35]

Answer:

$0.276

Step-by-step explanation:

Given data

Bill= $9.20

Tax= 3%

Tip= 19%

Let us find the amount of the tax and the tip

Tax

=3/100*9.20

=0.03*9.2

=$0.276

Amount after sales tax

= 0.276+9.20

=$9.476

Tip

=19/100*9.476

=0.19*9.476

=$1.80044

Therefore the sales tax

=$0.276

4 0
3 years ago
£90 is divided between Mark, Henry &amp; Gavin so that Mark gets twice as much as Henry, and Henry gets three times as much as G
Phantasy [73]

Answer:

Mark gets £54

Step-by-step explanation:

Let Gavin share=x

Henry=3x

Mark=2(3x)

x+3x+2(3x)=£90

x+3x+6x=£90

10x=£90

x=9

Gavin=x=£9

Henry=3x

=3(9)=£27

Mark=2(3x)

=2(3*9)

=2(27)

=£54

4 0
3 years ago
Dr. Miriam Johnson has been teaching accounting for over 20 years. From her experience, she knows that 60% of her students do ho
oksano4ka [1.4K]

Answer:

a) The probability that a student will do homework regularly and also pass the course = P(H n P) = 0.57

b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P') = 0.12

c) The two events, pass the course and do homework regularly, aren't mutually exclusive. Check Explanation for reasons why.

d) The two events, pass the course and do homework regularly, aren't independent. Check Explanation for reasons why.

Step-by-step explanation:

Let the event that a student does homework regularly be H.

The event that a student passes the course be P.

- 60% of her students do homework regularly

P(H) = 60% = 0.60

- 95% of the students who do their homework regularly generally pass the course

P(P|H) = 95% = 0.95

- She also knows that 85% of her students pass the course.

P(P) = 85% = 0.85

a) The probability that a student will do homework regularly and also pass the course = P(H n P)

The conditional probability of A occurring given that B has occurred, P(A|B), is given as

P(A|B) = P(A n B) ÷ P(B)

And we can write that

P(A n B) = P(A|B) × P(B)

Hence,

P(H n P) = P(P n H) = P(P|H) × P(H) = 0.95 × 0.60 = 0.57

b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P')

From Sets Theory,

P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1

P(H n P) = 0.57 (from (a))

Note also that

P(H) = P(H n P') + P(H n P) (since the events P and P' are mutually exclusive)

0.60 = P(H n P') + 0.57

P(H n P') = 0.60 - 0.57

Also

P(P) = P(H' n P) + P(H n P) (since the events H and H' are mutually exclusive)

0.85 = P(H' n P) + 0.57

P(H' n P) = 0.85 - 0.57 = 0.28

So,

P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1

Becomes

0.03 + 0.28 + 0.57 + P(H' n P') = 1

P(H' n P') = 1 - 0.03 - 0.57 - 0.28 = 0.12

c) Are the events "pass the course" and "do homework regularly" mutually exclusive? Explain.

Two events are said to be mutually exclusive if the two events cannot take place at the same time. The mathematical statement used to confirm the mutual exclusivity of two events A and B is that if A and B are mutually exclusive,

P(A n B) = 0.

But, P(H n P) has been calculated to be 0.57, P(H n P) = 0.57 ≠ 0.

Hence, the two events aren't mutually exclusive.

d. Are the events "pass the course" and "do homework regularly" independent? Explain

Two events are said to be independent of the probabilty of one occurring dowant depend on the probability of the other one occurring. It sis proven mathematically that two events A and B are independent when

P(A|B) = P(A)

P(B|A) = P(B)

P(A n B) = P(A) × P(B)

To check if the events pass the course and do homework regularly are mutually exclusive now.

P(P|H) = 0.95

P(P) = 0.85

P(H|P) = P(P n H) ÷ P(P) = 0.57 ÷ 0.85 = 0.671

P(H) = 0.60

P(H n P) = P(P n H)

P(P|H) = 0.95 ≠ 0.85 = P(P)

P(H|P) = 0.671 ≠ 0.60 = P(H)

P(P)×P(H) = 0.85 × 0.60 = 0.51 ≠ 0.57 = P(P n H)

None of the conditions is satisfied, hence, we can conclude that the two events are not independent.

Hope this Helps!!!

7 0
3 years ago
671 ÷24showing work please and thank you
telo118 [61]

Answer: 27.958333333333333333333333333333........

Step-by-step explanation: Use long division method or do rounding method round each number and divide.

6 0
3 years ago
Read 2 more answers
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