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Mashutka [201]
3 years ago
7

The answer the the question pictured

Mathematics
1 answer:
stiks02 [169]3 years ago
6 0

Answer:

x-1/2x-3

Step-by-step explanation:

be happy bro this is your perfect answer

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Solve m -5m+18=-m-18
zubka84 [21]
When I solved it the answer I got was m=9
6 0
3 years ago
Write the first 4 terms for this sequence t(n)=4.5n-8
siniylev [52]
Evaluate t(1),t(2),t(3),t(4)

t(1)=4.5(1)-8=4.5-8=-3.6
t(2)=4.5(2)-8=9-8=1
t(3)=4.5(3)-8=13.5-8=5.5
t(4)=4.5(4)-8=18-8=10

the first 4 terms are
-3.6,1,5.5, 10
6 0
4 years ago
Read 2 more answers
7.) Solve the system of equations<br> below. <br> -2x + 3y = 9 2/3x + 2y = 0
iren [92.7K]

Answer:

So lets isolate y

we have

-2x+3y=9

2/3x +2y=0

Im gonna do the first equation

so -2x + 3y = 9

Add 2x on both sides

3y= 9+ 2x

Now divide by 3 to isolate y

y= 3+ 2/3x

Now we can subsitute y into any eqaution (I’m gonna do the second one) to solve.

2/3x+ 2(3+ 2/3x)= 0

Solve:

2/3x+ 6 + 4/3x = 0

6/3x +6= 0

2x+6=0

2x= -4

x= -2

Now we can subsitute this for any equation.

so I’m gonna do the first equation

-2(-2)+3y=9

4+3y=9

3y= 5

y= 5/3

y= 1 2/3

Final answer:

x=-2

y= 1 2/3

6 0
2 years ago
What are the first four terms of the sequence modeled by the recursive formula a n =a n-1 +5, when a1 = -3?
Brut [27]

Answer:2,7,12,19

Step-by-step explanation:

a2=a2-1+5=-3+5=2

a3=a3-1+5=a2+5=2+5=7

a4=a3+5=12 and continue so on

3 0
3 years ago
Read 2 more answers
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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