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Diameter Radius Area Circumference of= 132

AveGali [126]

Answer:

$lets \: \: find \: \: \: the \: \: \: radius \\ 2\pi \: r = 132 \\ \frac{2\pi \: r}{2\pi} = \frac{132}{2\pi} \\ r = 21.01$

$lets \: \: \: find \: \: \: the \:area \: now \\ = \pi {r}^{2} \\ = \pi \times {21.01}^{2} \\ = 1,386.76$

$lets \: fin d\: \: \: the \: \: diameter \\ = d = 2r \\ = d = 2 \times 21.01 \\ = d = 42.02$

7 0

3 years ago

Given f(x) = -x – 4, find f(4). I need help

Brums [2.3K]

Answer:

f(4) = -8

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Function Notation

Step-by-step explanation:

Step 1: Define

f(x) = -x - 4

f(4) is x = 4

Step 2: Evaluate

Substitute in x: f(4) = -4 - 4
Subtract: f(4) = -8

5 0

3 years ago

A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature

Westkost [7]

Answer:

(a) The value of k is $\frac{1}{15}$ .

(b) The probability that at most three forms are required is 0.40.

(c) The probability that between two and four forms (inclusive) are required is 0.60.

(d) $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of y.

Step-by-step explanation:

The random variable Y is defined as the number of forms required of the next applicant.

The probability mass function is defined as:

$P(y) = \left \{ {{ky};\ for \ y=1,2,...5 \atop {0};\ otherwise} \right$

(a)

The sum of all probabilities of an event is 1.

Use this law to compute the value of k.

$\sum P(y) = 1\\k+2k+3k+4k+5k=1\\15k=1\\k=\frac{1}{15}$

Thus, the value of k is $\frac{1}{15}$ .

(b)

Compute the value of P (Y ≤ 3) as follows:

$P(Y\leq 3)=P(Y=1)+P(Y=2)+P(Y=3)\\=\frac{1}{15}+\frac{2}{15}+ \frac{3}{15}\\=\frac{1+2+3}{15}\\ =\frac{6}{15} \\=0.40$

Thus, the probability that at most three forms are required is 0.40.

(c)

Compute the value of P (2 ≤ Y ≤ 4) as follows:

$P(2\leq Y\leq 4)=P(Y=2)+P(Y=3)+P(Y=4)\\=\frac{2}{15}+\frac{3}{15}+\frac{4}{15}\\ =\frac{2+3+4}{15}\\ =\frac{9}{15} \\=0.60$

Thus, the probability that between two and four forms (inclusive) are required is 0.60.

(d)

Now, for $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ to be the pmf of Y it has to satisfy the conditions:

$P(y)=\frac{y^{2}}{50}>0;\ for\ all\ values\ of\ y \\$
$\sum P(y)=1$

Check condition 1:

$y=1:\ P(y)=\frac{y^{2}}{50}=\frac{1}{50}=0.02>0\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.08>0 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.18>0\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.32>0 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.50>0$

Condition 1 is fulfilled.

Check condition 2:

$\sum P(y)=0.02+0.08+0.18+0.32+0.50=1.1>1$

Condition 2 is not satisfied.

Thus, $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of y.

7 0

3 years ago

Test pls help! thank you

natita [175]

Answer:

The Answer is 5/12

Pls mark as brainliest

5 0

3 years ago

Read 2 more answers

the diagram shows a right angles triangle ABC. BC = 10 cm. angle CAB = 90 degrees. angle ABC = 30 degrees. M is the midpoint of

Karolina [17]

Answer:

angle AMC=49.1

Step-by-step explanation:

8 0

3 years ago