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Fynjy0 [20]
3 years ago
11

A cars tire pressure decreased by 0.098 of its original pressure write 0.098 as a percent

Mathematics
2 answers:
Anastasy [175]3 years ago
8 0
Answer would be 9.8% mate
balandron [24]3 years ago
5 0
To write this as a percent, just move the decimal two places to the right. So instead of .098, this becomes 9.8%. hope this helps!
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Using the law of sins:

B =sin^−1((sin(A)⋅a)/b) = sin^−1((sin(65)⋅110)/155)=40.03 degrees

B= 40.03 degrees (round as needed)

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2
kozerog [31]

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Step-by-step explanation:

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Jimmy scored a total of 80 points on 2 tests. He scored 10 points more on the 2nd test. What was his score on the first test?
zavuch27 [327]
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5 0
4 years ago
(a) How many integers from 1 through 1,000 are multiples of 2 or multiples of 9?
DENIUS [597]

Answer:

(a)There are 500 integers from 1 through 1,000 are multiples of 2.

There are 111 integers from 1 through 1,000 are multiples of 9.

(b)0.611

Step-by-step explanation:

Given the set of Integers from 1 through 1000

The least multiple of 2 is 2 and the highest multiple of 2 in the interval is 1000.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 2,4,6,... is an Arithmetic Progression,

where first term, a =2, common difference, d =2

Nth term of an A.P,

U_n= a+(n-1)d\\1000=2+2(n-1)\\1000=2+2n-2\\1000=2n\\n=500

  • There are 500 integers from 1 through 1,000 are multiples of 2.

Similarly,

Given the set of Integers from 1 through 1000

The least multiple of 9 is 9 and the highest multiple of 9 in the interval is 999.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 9,18,27,... is an Arithmetic Progression,

where first term, a =9, common difference, d =9

Nth term of an A.P,

U_n= a+(n-1)d\\999=9+9(n-1)\\999=9+9n-9\\999=9n\\n=111

  • There are 111 integers from 1 through 1,000 are multiples of 9.

(b)Probability that an integer selected at random is a multiple of 2 or a multiple of 9.

There are a total of 1000 numbers between from 1 through 1,000

n(S)=1000

n(Multiples of 2)=500

n(Multiples of 9)=111

Therefore:

P(\text{Multiples of 9}) \:OR\: P(\text{Multiples of 2})=\frac{111}{1000} + \frac{500}{1000} \\=\frac{611}{1000} \\=0.611

7 0
3 years ago
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