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Mila [183]
3 years ago
13

6=2(y+2) what is the value of y in the equation

Mathematics
2 answers:
uysha [10]3 years ago
8 0

Answer: y = 1

Step-by-step explanation:

Leona [35]3 years ago
6 0

Answer: the value of y is 1

Step-by-step explanation: 6=2 x 1+2

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ahrayia [7]

Answi do not understand your question

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3 years ago
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Y=a|x-h|+k please help
RSB [31]

Answer:

i can't answer this but i can tell you this

The general form of an absolute value function is f(x)=a|x-h|+k. From this form, we can draw graphs.

y = a(x – h)2 + k, where (h, k) is the vertex. ... In the vertex form of the quadratic, the fact that (h, k) is the vertex makes sense if you think about it for a minute, and it's because the quantity "x – h" is squared, so its value is always zero or greater; being squared, it can never be negative.

Step-by-step explanation:

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7 0
2 years ago
In the equation -y = 10x, what is the unit rate?
Umnica [9.8K]
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7 0
3 years ago
The heights of men in a certain population follow a normal distribution with mean 69.7 inches and standard deviation 2.8 inches.
Mama L [17]

Answer:

a) P(Y > 76) = 0.0122

b) i) P(both of them will be more than 76 inches tall) = 0.00015

   ii) P(Y > 76) = 0.0007

Step-by-step explanation:

Given - The heights of men in a certain population follow a normal distribution with mean 69.7 inches and standard deviation 2.8 inches.

To find - (a) If a man is chosen at random from the population, find

                    the probability that he will be more than 76 inches tall.

              (b) If two men are chosen at random from the population, find

                    the probability that

                    (i) both of them will be more than 76 inches tall;

                    (ii) their mean height will be more than 76 inches.

Proof -

a)

P(Y > 76) = P(Y - mean > 76 - mean)

                 = P( \frac{( Y- mean)}{S.D}) > \frac{( 76- mean)}{S.D})

                 = P(Z >  \frac{( 76- mean)}{S.D})

                 = P(Z > \frac{76 - 69.7}{2.8})

                 = P(Z > 2.25)

                 = 1 - P(Z  ≤ 2.25)

                 = 0.0122

⇒P(Y > 76) = 0.0122

b)

(i)

P(both of them will be more than 76 inches tall) = (0.0122)²

                                                                           = 0.00015

⇒P(both of them will be more than 76 inches tall) = 0.00015

(ii)

Given that,

Mean = 69.7,

\frac{S.D}{\sqrt{N} } = 1.979899,

Now,

P(Y > 76) = P(Y - mean > 76 - mean)

                 = P( \frac{( Y- mean)}{\frac{S.D}{\sqrt{N} } })) > \frac{( 76- mean)}{\frac{S.D}{\sqrt{N} } })

                 = P(Z > \frac{( 76- mean)}{\frac{S.D}{\sqrt{N} } })

                 = P(Z > \frac{( 76- 69.7)}{1.979899 }))

                 = P(Z > 3.182)

                 = 1 - P(Z ≤ 3.182)

                 = 0.0007

⇒P(Y > 76) = 0.0007

6 0
3 years ago
Given: △ACM, m∠C=90°, CP ⊥ AM
larisa86 [58]

Answer:  The answer is 3\dfrac{4}{7}.

Step-by-step explanation:  Given in the question that ΔAM is a right-angled triangle, where ∠C = 90°, CP ⊥ AM, AC : CM = 3 : 4 and MP - AP = 1. We are to find AM.

Let, AC = 3x and CM = 4x.

In the right-angled triangle ACM, we have

AM^2=AC^2+CM^2=(3x)^2+(4x)^2=9x^2+16x^2=25x^2\\\\\Rightarrow AM=5x.

Now,

AM=AP+PM=AP+(AP+1)\\\\\Rightarrow 2AP=AM-1\\\\\Rightarrow 2AP=5x-1.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(A)

Now, since CP ⊥ AM, so ΔACP and ΔMCP are both right-angled triangles.

So,

CP^2=AC^2-AP^2=CM^2-MP^2\\\\\Rightarrow (3x)^2-AP^2=(4x)^2-(AP+1)^2\\\\\Rightarrow 9x^2-AP^2=16x^2-AP^2-2AP-1\\\\\Rightarrow 2AP=7x^2-1.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(B)

Comparing equations (A) and (B), we have

5x-1=7x^2-1\\\\\Rightarrow 5x=7x^2\\\\\Rightarrow x=\dfrac{5}{7},~\textup{since }x\neq 0.

Thus,

AM=5\times\dfrac{5}{7}=\dfrac{25}{7}=3\dfrac{4}{7}.

8 0
4 years ago
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