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MrRa [10]
3 years ago
10

Um....help...please id.k where to start and says i need to show work so that's fun.

Mathematics
1 answer:
Lina20 [59]3 years ago
7 0

Answer:

2x/2x+3

Step-by-step explanation:

2x(3x+2) over 6x 2squared + 13 x 6 = 2x (3x+2) over (3x+2) (2x+3)

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The price of a ticket to the circus this year is 10% more than the price of a ticket to the circus last year. If the variable t
dlinn [17]

Answer:

Option B multiply t with 1.1

Step-by-step explanation:

Let

t------> the ticket price for the circus last year

x-----> the ticket price for the circus this year

we know that

100\%+10\%=110\%=110/100=1.10

so

x=1.10(t)

therefore

The ticket price for the circus this year is equal to the ticket price for the circus last year multiplied by 1.10

7 0
3 years ago
If the radius of a half-circle is 11...
Xelga [282]
By giving just the radius and asking for the arc length of that circle your answer will be without math is 22
π
feet

But;

Let’s do same math!

Arc length = (()/360)⋅2⋅⋅
(
(
a
n
g
l
e
o
f
a
r
c
)
/
360
)
⋅
2
⋅
π
⋅
r

In this case angle of arc was not specified so: 360∘
∘
was used.

Arc length = ((360)/360)⋅2⋅⋅11
(
(
360
)
/
360
)
⋅
2
⋅
π
⋅
11

Arc length = 22
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feet
5 0
3 years ago
Write the sum without sigma notation. then evaluate. summation from k equals 1 to 5 cosine k pi
erastova [34]

summation from k equals 1 to 5 cosine k pi

= Cosine 1*∏ + Cosine 2*∏ + Cosine 3*∏ + Cosine 4*∏ + Cosine 5*∏

= -1 + 1 + -1 + 1 + -1

= -1

5 0
3 years ago
3 $15.60 to $11.70X ХFind each percent change. (Increase or Decrease?). Round to thenearest whole percent.Example: 25% increase
Ksivusya [100]

The percentage change from $15.60 to $11.70 is computed as follows:

undefined

6 0
1 year ago
Hdc produces microcomputer hard drives at four different production facilities (f1, f2, f3, and f4) hard drive production at f1,
OLEGan [10]
Let D denote the event that an HD is defective, and F_i the event that a particular HD was produced at facility i.

You are asked to compute

\mathbb P(F_2\mid D)
\mathbb P(F_4\mid D)
\mathbb P(D^C)

From the definition of conditional probabilities, the first two will require that you first find \mathbb P(D). Once you have this, part (c) is trivial.

I'll demonstrate the computation for part (a). Part (b) is nearly identical.

(a)
\mathbb P(F_2\mid D)=\dfrac{\mathbb P(F_2\cap D)}{\mathbb P(D)}

Presumably, the facility responsible for producing a given HD is independent of whether the HD is defective or not, so \mathbb P(F_2\cap D)=\mathbb P(F_2)\mathbb P(D)=0.20\times0.015.

Use the law of total probability to determine the value of the denominator:

\mathbb P(D)=\mathbb P(D\mid F_1)+\mathbb P(D\mid F_2)+\mathbb P(D\mid F_3)+\mathbb P(D\mid F_4)

We know each of the component probabilities because they are given explicitly: 0.015, 0.02, 0.01, and 0.03, respectively. So

\mathbb P(D)=0.015+0.02+0.01+0.03=0.075

and thus

\mathbb P(F_2\mid D)=\dfrac{0.2\times0.015}{0.075}=0.04

(b) Similarly,
\mathbb P(F_4\mid D)=\dfrac{0.4\times0.03}{0.075}=0.16

(c)
\mathbb P(D^C)=1-\mathbb P(D)=0.925
6 0
3 years ago
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