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2 years ago

W+19=49 for w=30 help me

Mathematics

2 answers:

mafiozo [28]2 years ago

8 0

W + 19 = 49 for w = 30 :

30 + 19 = 49

hope this helps !.

mihalych1998 [28]2 years ago

3 0

W+19=49 w=30
30+19=49

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There are 10 cranberry and 15 strawberry juices. You randomly pick two juices. What's the probability you'll get two strawberry

lorasvet [3.4K]

Answer:

7/20

Explanation:

The theoretical probability, P (A), of an event, A, is:

P(A) = number of outcomes of event A / total number of possible outcomes

For 10 cranberry and 15 strawberry juices, the number of outcomes for two strawberry juices are:

Combination of 15 strawberry juices chosen in two:

$C^{15}_{2}=\frac{15!}{(15-2)!(2!)}=\frac{15!}{13!2!}=\frac{15\times 14}{2}=105$

The total number of possible outcomes is:

Combination of 25 juices chosen in two:

$C^{25}_{2}=\frac{25!}{(25-2)!(2!)}=\frac{25!}{23!2!}=\frac{25\times 24}{2}=300$

Thus, the probability of randomly picking two strawberry juices is:

P (two strawberry juices) = 105/300 = 7/20

Note that you can obtain it as the product of the probabilities that the first juice is a strawberry juice and the second juice is also strawberry juice:

$15/25\times 14/24=(15\times 25)/(14\times 24)=210/600=7/20$

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2 years ago

If the points (–2, –1), (–1, 0), (2, 3), and (–3, –2) are joined to form a straight line, at what point does the line intersect

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Step-by-step explanation:

Pick 2 points and use the slope formula

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Answer:

3y+3

Step-by-step explanation:

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2 years ago

The surface area of a right circular cone of radius r and height h is S = πr√ r 2 + h 2 , and its volume is V = 1 3 πr2h. What i

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Answer:

Required largest volume is 0.407114 unit.

Step-by-step explanation:

Given surface area of a right circular cone of radious r and height h is,

$S=\pi r\sqrt{r^2+h^2}$

and volume,

$V=\frac{1}{3}\pi r^2 h$

To find the largest volume if the surface area is S=8 (say), then applying Lagranges multipliers,

$f(r,h)=\frac{1}{3}\pi r^2 h$

subject to,

$g(r,h)=\pi r\sqrt{r^2+h^2}=8\hfill (1)$

We know for maximum volume $r\neq 0$ . So let $\lambda$ be the Lagranges multipliers be such that,

$f_r=\lambda g_r$

$\implies \frac{2}{3}\pi r h=\lambda (\pi \sqrt{r^2+h^2}+\frac{\pi r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}r h= \lambda (\sqrt{r^2+h^2}+\frac{ r^2}{\sqrt{r^2+h^2}})\hfill (2)$

And,

$f_h=\lambda g_h$

$\implies \frac{1}{3}\pi r^2=\lambda \frac{\pi rh}{\sqrt{r^2+h^2}}$

$\implies \lambda=\frac{r\sqrt{r^2+h^2}}{3h}\hfill (3)$

Substitute (3) in (2) we get,

$\frac{2}{3}rh=\frac{r\sqrt{R^2+h^2}}{3h}(\sqrt{R^2+h^2+}+\frac{r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}rh=\frac{r}{3h}(2r^2+h^2)$

$\implies h^2=2r^2$

Substitute this value in (1) we get,

$\pi r\sqrt{h^2+r^2}=8$

$\implies \pi r \sqrt{2r^2+r^2}=8$

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Then,

$h=\sqrt{2}(1.21252)\equiv 1.71476$

Hence largest volume,

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