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emmainna [20.7K]

3 years ago

11

Solve for X. I need help

1 answer:

Nuetrik [128]3 years ago

5 0

Answer:

im just tryna get points

Step-by-step explanation:

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Plz help 100 points

tekilochka [14]

Answer:

the answer is one line is increased by 1, and the other increased by 6

hope this helped :))

4 0

3 years ago

Read 2 more answers

In a study, 36% of adults questioned reported that their health was excellent. A researcher wishes to study the health of peop

steposvetlana [31]

Answer:

0.3907

Step-by-step explanation:

We are given that 36% of adults questioned reported that their health was excellent.

Probability of good health = 0.36

Among 11 adults randomly selected from this area, only 3 reported that their health was excellent.

Now we are supposed to find the probability that when 11 adults are randomly selected, 3 or fewer are in excellent health.

i.e. $P(x\leq 3)=P(x=1)+{P(x=2)+P(x=3)$

Formula : $P(x=r)=^nC_r p^r q ^ {n-r}$

p is the probability of success i.e. p = 0.36

q = probability of failure = 1- 0.36 = 0.64

n = 11

So, $P(x\leq 3)=P(x=1)+{P(x=2)+P(x=3)$

$P(x\leq 3)=^{11}C_1 (0.36)^1 (0.64)^{11-1}+^{11}C_2 (0.36)^2 (0.64)^{11-2}+^{11}C_3 (0.36)^3 (0.64)^{11-3}$

$P(x\leq 3)=\frac{11!}{1!(11-1)!} (0.36)^1 (0.64)^{11-1}+\frac{11!}{2!(11-2)!} (0.36)^2 (0.64)^{11-2}+\frac{11!}{3!(11-3)!} (0.36)^3 (0.64)^{11-3}$

$P(x\leq 3)=0.390748$

Hence the probability that when 11 adults are randomly selected, 3 or fewer are in excellent health is 0.3907

5 0

3 years ago

What's the answer please help

Talja [164]

$\frac{55}{550} * \frac{100}{1} =x\\
\frac{55*100}{550*1}=x\\
\frac{5500}{550}=x\\
x=10$

3 0

3 years ago

Read 2 more answers

For each of the following vector fields

olga nikolaevna [1]

(A)

$\dfrac{\partial f}{\partial x}=-16x+2y$

$\implies f(x,y)=-8x^2+2xy+g(y)$

$\implies\dfrac{\partial f}{\partial y}=2x+\dfrac{\mathrm dg}{\mathrm dy}=2x+10y$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=10y$

$\implies g(y)=5y^2+C$

$\implies f(x,y)=\boxed{-8x^2+2xy+5y^2+C}$

(B)

$\dfrac{\partial f}{\partial x}=-8y$

$\implies f(x,y)=-8xy+g(y)$

$\implies\dfrac{\partial f}{\partial y}=-8x+\dfrac{\mathrm dg}{\mathrm dy}=-7x$

$\implies \dfrac{\mathrm dg}{\mathrm dy}=x$

But we assume $g(y)$ is a function of $y$ alone, so there is not potential function here.

(C)

$\dfrac{\partial f}{\partial x}=-8\sin y$

$\implies f(x,y)=-8x\sin y+g(x,y)$

$\implies\dfrac{\partial f}{\partial y}=-8x\cos y+\dfrac{\mathrm dg}{\mathrm dy}=4y-8x\cos y$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=4y$

$\implies g(y)=2y^2+C$

$\implies f(x,y)=\boxed{-8x\sin y+2y^2+C}$

For (A) and (C), we have $f(0,0)=0$ , which makes $C=0$ for both.

4 0

3 years ago

Hi what is 4x3-1 x20=<br><br><br><br><br><br> Don’t mind this

krek1111 [17]

Answer:

4 x 3 - 1 x 20 = -8

H u m a n F i n g e r P i e.

3 0

3 years ago

Read 2 more answers