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Anarel [89]
3 years ago
6

Using your calculator, find the proportion of observations from a standard normal distribution that satisfies the following stat

ement. Sketch the normal curve and shade the area under the curve that is the answer to the question: Z < –1.5.
Mathematics
1 answer:
emmainna [20.7K]3 years ago
3 0
<span>Look at your table for a Z value of 1.55. The numbers on the far left column are your z values. See the 1.5 row, then move over to the 0.05 column to make it 1.55.
You'll see 0.9394.
That's the area under the normal curve from 1.55 to negative infinity.
But you wanted the area under the curve greater than 1.55.
Take 1-0.9394=0.0606.
You subtract from 1 because you know that the area under the whole curve is 1, so it gives you the area you need.</span>
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F(2) = -13 and f(-3) = 12
mario62 [17]

Answer:

f(x) = -5x-3

Step-by-step explanation:

f(x) = ax + b

f(2)=-13 & f(-3) =12

2a + b = -13

-3a+ b = 12

_________-

5a = -25

a = -5

b= -13+10= -3

so the formula of f(x) = -5x -3

5 0
3 years ago
Which of the following is an extraneous solution of (45-3x)^((1)/(2))=x-9
Anna007 [38]

Answer:

c x =3

Step-by-step explanation:

5 0
3 years ago
In the multiplication sentence below, which numbers are the factors? Check
kompoz [17]
B and c are the factors because they make the product
5 0
3 years ago
Read 2 more answers
1 pt) If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −4≤u≤4,−4≤v≤4, has surface area equal to 1, what is t
natta225 [31]

The area of the surface given by \vec r_1(u,v) is 1. In terms of a surface integral, we have

1=\displaystyle\int_{-4}^4\int_{-4}^4\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv

By multiplying each component in \vec r_1 by 5, we have

\dfrac{\partial\vec r_2(u,v)}{\partial u}=5\dfrac{\partial\vec r_1(u,v)}{\partial u}

and the same goes for the derivative with respect to v. Then the area of the surface given by \vec r_2(u,v) is

\displaystyle\int_{-4}^4\int_{-4}^425\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\boxed{25}

8 0
3 years ago
Solve the equation 58 - 10x ≤ 20 + 9x
drek231 [11]

Answer:

x≥2

Step-by-step explanation:

First, write out the equation as you have it:

58-10x\leq20+9x

Then, add 10x to both sides:

58-10x+10x\leq20+9x+10x\\58\leq20+19x

Next, subtract 20 from both sides:

58-20\leq20-20+19x\\38\leq19x

Finally, divide both sides by 19:

\frac{38}{19}\leq\frac{19x}{19}\\2\leq x

or

x\geq 2

Therefore: the answer to this inequality/equation is: x≥2

6 0
3 years ago
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