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Nonamiya [84]
3 years ago
12

There's 2 questions here that need to be answered. Find the value of each variable.

Mathematics
1 answer:
gladu [14]3 years ago
6 0

Answer:

5. x= 13 y= 18.4

6.

Step-by-step explanation:

question 5:

tan45=x/13

multiply both sides by 13 to get x by its self.

end up with 13×tan45 so put that in the calculator to get x

then you do cos45= 13/y

multiply by y on both sides.

end up with y×cos45=13

divide the cos45 by both sides

put 13/cos45 to get y

question 6:

sin30=4/x

multiply x on both sides

end with x×sin30 =4

divide sin30 on both sides

put 4/sin30 in a calculator to get x

same thing again but with tan30 ahaha

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balandron [24]
Hey, here’s your answer:)

4 0
2 years ago
a water ballon is to be randomly dropped along a 1 mile long line that stretches across a practice target zone. the target zones
Pavel [41]

Answer:

It seems like the question is not complete. So, I will asume that the complete question is: " A bomb is to be dropped along a mile-long line that stretches across a practice target.  The target center is at the midpoint of the line.  The target will be destroyed if the bomb falls within a tenth of a mile on either side of the center.  Find the Probability that the target is destroyed if the bomb falls randomly along the line."

Step-by-step explanation:

The total of possible cases is the length of the line = 1 mi ;

The favourable cases are the two lengths of 0.1 mi = 0.2 mi ;

Assuming the bomb has no bias for any point ,

the probability of favourable cases' occurrence is 0.2/1 = 0.2

7 0
2 years ago
The Transportation Safety Authority (TSA) has developed a new test to detect large amounts of liquid in luggage bags.
Ivan

Answer and Step-by-step explanation:

First of all lets define  

X to be that the bags do contain large liquid amounts

+ to be that the test is not negative that is positive

From the question,  

P(x) = 3/100 = 0.03

P(+|X) = 0.91

P(+|X′) = 0.05

Probability of bag having a positive test =p(+)

= P(+|X) (P(x)) + P(+|X′)(P(X′))

= P(X′)) = 1 – 0.03 = 0.97

Inserting these values into these formulas  

0.91)(0,03) + (0.05)(0.97)

= 0.0273 + 0.0485

= 0.0758

The probability of the randomly selected bag having large liqid amount

=  P(X|+) = (0.91*0.03)/(0.91*0.03)(0.05*0.97)

= 0.0273/0.0273+0.0485

= 0.0273/0.0758

= 0.3602

the probability that this bag does not have large liquid amount

= p(X'|+) = 1 - P(X|+)

= 1 - 0.3602

= 0.6398

6 0
3 years ago
Ann bought 2 pounds of cheese to make lasagna the recipe gived the amount of cheese needed in onces How mojaunces how many ounce
allochka39001 [22]
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2 years ago
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Vinvika [58]

Answer:

D

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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