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3 years ago

Find the z-score such as 88% of the area under the normal curve is to the left of the z-score

1 answer:

6 0

If you have a calculator with statistical functions, that's the way to go.

On my TI-83, I typed in invNorm(0.88) and got the result z = 1.17.

88% of the area under the normal curve is to the left of z = 1.17.

You might be interested in

Does anyone know how to do this?

sineoko [7]

Answer:

$\boxed{\tt x=\sqrt{85} }$

Step-by-step explanation:

When making a right triangle with 2 points, the distance is the hypotenuse which can be represent as "x"

To find the distance we can use the Pythagorean theorem.

$\boxed{{\tt a^2+b^2=c^2}}$

→ $\tt a=9$

→ $\tt b=2$

→ $\tt c=x$

Let's Plug in the side lengths:

$\tt 9^2+2^2=x^2$

Evaluate 9²= 81 and 2²= 4:

$\tt 81+4=x^2$

Now, add 81+4= 85

$\tt 85=x^2$

Take the square root of both sides:

$\tt x=\sqrt{85}$

_____________________________

5 0

2 years ago

if you want to buy a shirt that cost 18.99 but you have 20% off coupon and the tax is 7%. how much does the shirt cost

dusya [7]

20% off means you pay 80% of the price so you multiply by 0.8, and 7% tax means you multiply by 1.07

Resulting cost = 18.99*0.8*1.07 = 16.26

Hope this helps let me know if im wrong!

7 0

3 years ago

Read 2 more answers

Graph the hyperbola using the transverse axis, vertices, and co-vertices:

Reptile [31]

See attachment for the graph of the hyperbola 12x^2 - 3y^2 - 108 = 0

<h3>How to graph the hyperbola?</h3>

The equation of the hyperbola is given as:

12x^2 - 3y^2 - 108 = 0

Start by calculating the transverse axis

So, we have:

Transverse axis

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

Where

a = 3 and b = 6

The transverse axis is calculated as:

y = ±b/a(x - h) + k

So, we have:

y = ±6/3(x - 0) + 0

Evaluate the difference and sum

y = ±6/3x

Evaluate the quotient

y = ±2x

This means that the transverse axes are y = 2x and y =-2x

The vertices

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

The co-vertices

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

And

a = 3 and b = 6

The co-vertices are

(h - a, k) and (h + a, k)

So, we have:

(0 - 3, 0) and (0 + 3, 0)

Evaluate

(-3,0) and (3, 0)

See attachment for the graph of the hyperbola