У— пинь 7. Graph y= 4x-3

While Edward was visiting his sister in Westminster, he bought a toothbrush that was marked down 80% from an original price of $

Andre45 [30]

Answer:

$1.78

Step-by-step explanation:

5 0

3 years ago

Write a compound inequality for the graph shown below. Use x for your variable.

ICE Princess25 [194]

Answer:

wheres your graph?

Step-by-step explanation:

i could do this with a graph lol

3 0

3 years ago

Read 2 more answers

Enter an inequality that represents the description, and then solve. Dave has $14 to spend on a $7 book and two birthday cards (

iogann1982 [59]

Answer:

3.5

Step-by-step explanation:

14 - 7 = 7

7/2

3.5

so thats the most hope I helped

4 0

3 years ago

A cone with a radius of 3" has a total area of 24π sq. in. Find the volume of the cone.

Rus_ich [418]

The answer is 12π.

To get the volume of the cone, we need the height. The radius is given.

V = πr² × (h/3)

The total surface area of the cone is:
SA = πr² + πrl where r is radius and l is slant height

24π = π(3)² + π(3)(l)
24π = 9π + 3πl
24π - 9π = 3πl
15π = 3πl
l = 15π / 3π
l = 5

Using Phytagoras, we can calculate the height of the cone:
l² = h² + r²
5² = h² + 3²
25 - 9 = h²
h = √16
h = 4

Therefore the volume is:
V = π(3)² × (4 / 3)
V = 3π × 4
V = 12π

5 0

3 years ago

Read 2 more answers

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