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sineoko [7]
2 years ago
10

Officer Brimberry wrote 32 tickets for traffic violations last week, but only 12 tickets this week. What is the percent decrease

? Give your answer to the nearest tenth of a percent. [The percent decrease is___%​
Mathematics
1 answer:
kolbaska11 [484]2 years ago
4 0

Answer:

62.5 %

Step-by-step explanation:

you have to subtract 32-12, which equals 20

then divide 20 by the original amount (20/32)

20/32= .625, then you have to multiply to get the final answer...hope this helped!

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1
Nadusha1986 [10]

Answer: 9 × 6 = 54

Step-by-step explanation:

In the Product Game, we need to multiply the both number of factor markers.

Chris and Katie were playing The Product Game. Their factor markers were on 9  and 2.

Number = 9 × 2 = 18

Chris decided to move the marker from 2 to 6.

Now, one maker is on 9 and other is on 6. So,

Number = 9 × (2 + 4)

Number = 9 × 6 = 54

Therefore, 9 × 6 = 54 is a numerical expression  to represent his move.​

3 0
3 years ago
Find cotθ, cosθ, and secθ, where θ is the angle shown in the figure.
Darina [25.2K]

Answer:

\cos( \theta)  =  \frac{5}{8}  \\  {8}^{2}  =  {5}^{2}  +  {opp}^{2}  \\ {opp}^{2}  = 64 - 25 = 39 \\ opp =  \sqrt{39}  \\  \sin(\theta)  =  \frac{\sqrt{39}}{8}  \\  \tan(\theta)  =  \frac{ \sqrt{39} }{5}  \\  \cot(\theta)  =  \frac{1}{\tan(\theta)}  =  \frac{1}{\frac{ \sqrt{39} }{5}}  \\ \cot(\theta)  =  \frac{5}{ \sqrt{39} }  \\  \csc( \theta) =  \frac{1}{\sin(\theta)}  =  \frac{1}{\frac{\sqrt{39}}{8}}   \\ \csc( \theta) = \frac{8}{ \sqrt{39} }  \\  \sec( \theta) =  \frac{1}{\cos( \theta) }  =  \frac{1}{ \frac{5}{8} }  \\ \sec( \theta) =  \frac{8}{5}

5 0
2 years ago
use distributive property to simplify the expression 3 (8x + 2y) I need help with my algebra homework someone please help.
inysia [295]

6 \times (4 \times x + y)
4 0
3 years ago
How many solutions does this linear system have?
Andrews [41]

Answer:

x = 0 , y = 4

Step-by-step explanation:

Solve the following system:

{y = 4 - 3 x | (equation 1)

x + 2 y = 8 | (equation 2)

Express the system in standard form:

{3 x + y = 4 | (equation 1)

x + 2 y = 8 | (equation 2)

Subtract 1/3 × (equation 1) from equation 2:

{3 x + y = 4 | (equation 1)

0 x+(5 y)/3 = 20/3 | (equation 2)

Multiply equation 2 by 3/5:

{3 x + y = 4 | (equation 1)

0 x+y = 4 | (equation 2)

Subtract equation 2 from equation 1:

{3 x+0 y = 0 | (equation 1)

0 x+y = 4 | (equation 2)

Divide equation 1 by 3:

{x+0 y = 0 | (equation 1)

0 x+y = 4 | (equation 2)

Collect results:

Answer: {x = 0 , y = 4

8 0
2 years ago
A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as
Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let <em>X</em> = number of seconds in the batch.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.31.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

              =1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

                            ={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

8 0
3 years ago
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