I need help im too lazy still :)

The probability of drawing a silver ball out of a certain bag is

Lubov Fominskaja [6]

Answer:2/3

Step-by-step explanation:

7 0

3 years ago

The diameter of a sphere is 4 centimeters.

ELEN [110]

The volume of a sphere refers to the number of cubic units that will exactly fill a sphere. The volume of a sphere can be found or calculate by using the formula V=4/3πr^3, where r represents the radius of the figure.

In this exercise is given that a sphere has a radius of 4 centimeters and it is asked to find its volume and use 3.14 as the value of π or pi. The first step would be substitute the values into the previous mention formula.

V=4/3πr^3
V=4/3(3.14)(4 cm)^3
V=4/3)(3.14)(64 cm^3)
V=267.9 cm^3
The volume of the sphere is 267.9 cubic centimeters.

7 0

4 years ago

Read 2 more answers

What is the image of (6,-6) after a dilation by a scale factor of 4 centered at the origin

skelet666 [1.2K]

9514 1404 393

Answer:

(24, -24)

Step-by-step explanation:

Dilation centered at the origin multiplies each coordinate by the dilation factor.

image = 4 × (6, -6)

image = (24, -24)

6 0

3 years ago

A squirrel jumps along the number line. Each of its jumps is 5 units long. The squirrel sometimes jumps to the right and sometim

NeX [460]

See I would love to help but need the last part of the question. At what point?

7 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

3 years ago