Yall can u please help me this is a grade i really need All answers before 9:30

Si un cono lo obtienes haciendo girar un triángulo rectángulo cuya base es de 10cm y su altura es de 16cm ¿cuánto mide el radio

rodikova [14]

Answer:

10 cm.

Step-by-step explanation:

In order to better understand this question, I will attach an allusive image to the question.

Keep in mind that in this case the base of the triangle is equal to the radius of the base of the cone since the triangle rotates on its own axis, therefore the measurement of the base corresponds to 10 cm, therefore the radius measures 10 cm.

5 0

2 years ago

Pls help it’s due right now

gulaghasi [49]

Answer:

y + 11 = 14

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

What is the slope of the line that asses through these two points? (1, -2) (3, 4)

dusya [7]

$(\stackrel{x_1}{1}~,~\stackrel{y_1}{-2})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{4}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{4}-\stackrel{y1}{(-2)}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{1}}}\implies \cfrac{4+2}{2}\implies \cfrac{6}{2}\implies 3$

3 0

2 years ago

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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:

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