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3 years ago

5

Using the equation given identify the following:y=-x-9m= b=

1 answer:

lyudmila [28]3 years ago

6 0

Answer:

m = ___<u>−1</u>___ b = ___<u>−9</u>___

Step-by-step explanation:

In the Slope-Intercept Formula, y = mx + b, <em>b</em><em> </em>represents the y-intercept, and <em>m</em><em> </em>represents the <em>rate of</em><em> </em><em>change</em><em> </em>[<em>slope</em>], so with that being said, you have your answer.

I am joyous to assist you anytime.

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Help me idk what's the answer

Jobisdone [24]

76, 84, 93, 67, 82, 87, & 76

The mean is just the middle number of the data set,

Start by rewriting the set in order from smallest to biggest:
67, 76, 76, 82, 84, 87, 93

Determine which is the number in the middle,
67, 76, 76, *82*, 84, 87, 93

So, in this case, the median is 82.

4 0

2 years ago

Read 2 more answers

Tasha eats half her snack and gives the other half to two of her best friends to share equally. What portion of the whole snack

dybincka [34]

Answer:

tasha got 1/2 of it

one friend got 1/4 of it

the other friend got 1/4 of it

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

Suppose the sample space for a continuous random variable is 0 to 80. If the area under the density curve for the variable from

nikklg [1K]

Hello,

Area under the curve is 1.

So p(x>5)=1-p(x<5)=1-0.0625=0.9375

Answer A

4 0

2 years ago

The quotient of n and 12 subtracted from 100?

Korvikt [17]

Answer:

100 - 12/n

Step-by-step explanation:

quotient means division so

12/n

and subtracted from means say x - 12/n so

100 - 12/n

If you're looking to evaluate it then it would be

4(25n - 3) / n

5 0

2 years ago