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mrs_skeptik [129]
3 years ago
11

What is the perimeter of a rectangle with a length of 7 inches and a width of 6 2/3 inches?

Mathematics
1 answer:
Delvig [45]3 years ago
7 0
7x2=14
6 2/3x2=13 1/3
27 1/3
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A quality control worker checked that 3 out of 40 TVs are defective(do not work).
gregori [183]

Answer:

600

Step-by-step explanation:

Divide 8000 by 40 then multiply by 3

4 0
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Diagram 5 shows a triangle ABD such that AD = 17 cm, AB = 10 cm and BD = 21 cm. The straight line AC is perpendicular to the str
OlgaM077 [116]

Answer:

8

Step-by-step explanation:

100 = x^2 + AC^2

17^2 = AC^2 + (21 - x)^2

289 = AC^2 + 21^2 + x^2 - 2*21*x

289 =<u> AC^2</u> + 441 +<u> x^2</u> - 42x

from 1st equation AC^2 + x^2 = 100

289 = 441 + 100 - 42x

289 = 541 - 42x

42x = 541 - 289 = 252

x = 252/42 = 6

so AC^2 = 100 - 6^2 = 100 - 36 = 64

AC = 8

7 0
3 years ago
The area of a circular place is 45 square inches. Estimate the radius to the nearest integer.
aksik [14]
A=pi(r)^2
45=pi(r)^2   divide by pi (3.14)
14.33=r^2   square root both sides
3.79=r         round
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5 0
3 years ago
A college student is taking two courses. The probability she passes the first course is 0.73. The probability she passes the sec
zhenek [66]

Answer:

b) No, it's not independent.

c) 0.02

d) 0.59

e) 0.57

f) 0.5616

Step-by-step explanation:

To answer this problem, a Venn diagram should be useful. The diagram with the information of Event 1 and Event 2 is shown below (I already added the information for the intersection but we're going to see how to get that information in the b) part of the problem)

Let's call A the event that she passes the first course, then P(A)=.73

Let's call B the event that she passes the second course, then P(B)=.66

Then P(A∪B) is the probability that she passes the first or the second course (at least one of them) is the given probability. P(A∪B)=.98

b) Is the event she passes one course independent of the event that she passes the other course?

Two events are independent when P(A∩B) = P(A) * P(B)

So far, we don't know P(A∩B), but we do know that for all events, the next formula is true:

P(A∪B) = P(A) + P(B) - P(A∩B)

We are going to solve for P (A∩B)

.98 = .73 + .66 - P(A∩B)

P(A∩B) =.73 + .66 - .98

P(A∩B) = .41

Now we will see if the formula for independent events is true

P(A∩B) = P(A) x P(B)

.41 = .73 x .66

.41 ≠.4818

Therefore, these two events are not independent.

c) The probability she does not pass either course, is 1 - the probability that she passes either one of the courses (P(A∪B) = .98)

1 - P(A∪B) = 1 - .98 = .02

d) The probability she doesn't pass both courses is 1 - the probability that she passes both of the courses P(A∩B)

1 - P(A∩B) = 1 -.41 = .59

e) The probability she passes exactly one course would be the probability that she passes either course minus the probability that she passes both courses.

P(A∪B) - P(A∩B) = .98 - .41 = .57

f) Given that she passes the first course, the probability she passes the second would be a conditional probability P(B|A)

P(B|A) = P(A∩B) / P(A)

P(B|A) = .41 / .73 = .5616

4 0
4 years ago
Pls solve if u can &lt;3 much appreciated
Scorpion4ik [409]

Step-by-step explanation:

that is only possible, if we don't need to use a 2 and a 5 (but especially the 2) to create the 1 in the middle (log(10) = log(2×5) or log (5×2) = 1).

in other words, if the 1 in the middle is just a given, and does not use any of our digits for is creation, then a solution is possible.

first of all, the upper left corner has to be log(0⁰). we cannot use the digit 0 for anything else. and 0^n (except for n = 0) is not a valid argument for a logarithm.

log(0⁰) = 0 log(2×1) = 0.3... log(9/4) = 0.35...

log(1×2) = 0.3... 1 log(3×6) = 1.26...

log(7/3) = 0.37... log(4×5) = 1.3... log(6⁹) = 7.0...

but as soon as we need to build the center 1 by log(10), which takes then away one of the 2s (as the only way to build 10 by multiplication is 2×5 or 5×2), there is no possible solution.

7 0
2 years ago
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