Write an equation of a line perpendicular to line AB in slope-intercept form that passes through the point (7, 6).

Find the value of (-64)2/3

Marizza181 [45]

Answer:

-42.66666...

Step-by-step explanation:

-64 * 2/3 is -42.66666...

well, according to the calculator.

7 0

3 years ago

Read 2 more answers

The weight of an object on a particular scale is 145.2 lbs. The measured weight may vary from the actual weight by at most 0.3 l

joja [24]

1. Know what you're looking for:

The range is the difference between the lowest and highest values.

Your weight is 145.2 and can vary from the actual weight by maximum 0.3 lbs less or maximum 0.3 lbs more.

2. Calculate your lowest and highest values :

Lowest : 145.2 - 0.3 = 144.9 lbs

Highest : 145.2 + 0.3 = 145.5 lbs

3. Calculate the range of actual weights of the object :

145.5 - 144.9 = 0.6 lbs

--> The answer is: The range of actual weights of the object is 0.6 lbs.

There you go! I really hope this helped, if there's anything just let me know! :)

4 0

3 years ago

The line with the slope of -8 passing through (-11,7)

Dahasolnce [82]

Slope intercept form y = -8x - 81
Point slope form (y - 7) = -8 (x + 11)

5 0

3 years ago

20 POINTS Identify the type of transformation in the following graphic and describe the change.

o-na [289]

Answer:

Translation +6 Left

and maybe a slight dilation that makes it bigger.

Hope this helps!

6 0

3 years ago

Let z denote a random variable that has a standard normal distribution. Determine each of the probabilities below. (Round all an

Gelneren [198K]

Answer:

(a) P (Z < 2.36) = 0.9909 (b) P (Z > 2.36) = 0.0091

(c) P (Z < -1.22) = 0.1112 (d) P (1.13 < Z > 3.35) = 0.1288

(e) P (-0.77< Z > -0.55) = 0.0705 (f) P (Z > 3) = 0.0014

(g) P (Z > -3.28) = 0.9995 (h) P (Z < 4.98) = 0.9999.

Step-by-step explanation:

Let us consider a random variable, $X \sim N (\mu, \sigma^{2})$ , then $Z=\frac{X-\mu}{\sigma}$ , is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, $Z \sim N (0, 1)$ .

In statistics, a standardized score is the number of standard deviations an observation or data point is above the mean. The z-scores are standardized scores.

The distribution of these z-scores is known as the standard normal distribution.

(a)

Compute the value of P (Z < 2.36) as follows:

P (Z < 2.36) = 0.99086

≈ 0.9909

Thus, the value of P (Z < 2.36) is 0.9909.

(b)

Compute the value of P (Z > 2.36) as follows:

P (Z > 2.36) = 1 - P (Z < 2.36)

= 1 - 0.99086

= 0.00914

≈ 0.0091

Thus, the value of P (Z > 2.36) is 0.0091.

(c)

Compute the value of P (Z < -1.22) as follows:

P (Z < -1.22) = 0.11123

≈ 0.1112

Thus, the value of P (Z < -1.22) is 0.1112.

(d)

Compute the value of P (1.13 < Z > 3.35) as follows:

P (1.13 < Z > 3.35) = P (Z < 3.35) - P (Z < 1.13)

= 0.99960 - 0.87076

= 0.12884

≈ 0.1288

Thus, the value of P (1.13 < Z > 3.35) is 0.1288.

(e)

Compute the value of P (-0.77< Z > -0.55) as follows:

P (-0.77< Z > -0.55) = P (Z < -0.55) - P (Z < -0.77)

= 0.29116 - 0.22065

= 0.07051

≈ 0.0705

Thus, the value of P (-0.77< Z > -0.55) is 0.0705.

(f)

Compute the value of P (Z > 3) as follows:

P (Z > 3) = 1 - P (Z < 3)

= 1 - 0.99865

= 0.00135

≈ 0.0014

Thus, the value of P (Z > 3) is 0.0014.

(g)

Compute the value of P (Z > -3.28) as follows:

P (Z > -3.28) = P (Z < 3.28)

= 0.99948

≈ 0.9995

Thus, the value of P (Z > -3.28) is 0.9995.

(h)

Compute the value of P (Z < 4.98) as follows:

P (Z < 4.98) = 0.99999

≈ 0.9999

Thus, the value of P (Z < 4.98) is 0.9999.

**Use the z-table for the probabilities.

3 0

2 years ago