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adoni [48]
3 years ago
12

Which number is a solution of the inequality

Mathematics
1 answer:
SashulF [63]3 years ago
8 0

x(7-x)-8>8-8

x(7-x)-8>0


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A variety pack of snacks contains 4 pretzel bags, 3 popcorn bags and 3 cheddar cracker bags. What is the probability that someon
Svetlanka [38]

Answer:

2/15

Step-by-step explanation:

Given that :

pretzel = 4

Popcorn = 3

Cheddar cracker = 3

Total = 4 + 3 + 3 = 10

Probability = Required outcome / Total possible outcomes

Selecting with replacement :

First selection :

P(pretzel bag) = 4 / 10

Second selection :

P(pretzel bag) = 3 / 9

P(2 pretzel bags) = 4/10 * 3/9 = 12 / 90 = 2/15

5 0
3 years ago
Piper has a balance of $1,500 on
saul85 [17]

Answer: $408.96

Step-by-step explanation:

4 0
3 years ago
Suppose a simple random sample of size nequals36 is obtained from a population with mu equals 74 and sigma equals 6. ​(a) Descri
OlgaM077 [116]

Part a)

The simple random sample of size n=36 is obtained from a population with

\mu = 74

and

\sigma = 6

The sampling distribution of the sample means has a mean that is equal to mean of the population the sample has been drawn from.

Therefore the sampling distribution has a mean of

\mu = 74

The standard error of the means becomes the standard deviation of the sampling distribution.

\sigma_ { \bar X }  =  \frac{ \sigma}{ \sqrt{n} }  \\ \sigma_ { \bar X }  =  \frac{ 6}{ \sqrt{36} }  = 1

Part b) We want to find

P(\bar X \:>\:75.9)

We need to convert to z-score.

P(\bar X \:>\:75.9)  = P(z \:>\: \frac{75.9 - 74}{1} )  \\  = P(z \:>\: \frac{75.9 - 74}{1} ) \\  = P(z \:>\: 1.9) \\  = 0.0287

Part c)

We want to find

P(\bar X \: < \:71.95)

We convert to z-score and use the normal distribution table to find the corresponding area.

P(\bar X \: < \:71.95)  = P(z \: < \: \frac{71.9 5- 74}{1} )  \\  = P(z \: < \: \frac{71.9 5- 74}{1} ) \\  = P(z \: < \:  - 2.05) \\  = 0.0202

Part d)

We want to find :

P(73\:

We convert to z-scores and again use the standard normal distribution table.

P( \frac{73 - 74}{1} \:< \: z

5 0
3 years ago
Simplify the following.<br> -(3z)2<br> A. 6z<br> B. 9z2<br> C. -9z2<br> D. -6z2<br> E. 6z2
diamong [38]

Answer:

a or b

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
Problem 4: Solve the initial value problem
pishuonlain [190]

Separate the variables:

y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx

Separate the left side into partial fractions. We want coefficients a and b such that

\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}

\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}

\implies 1 = a(y-2)+b(y+1)

\implies 1 = (a+b)y - 2a+b

\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13

So we have

\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx

Integrating both sides yields

\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx

\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C

\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C

\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C

\dfrac{y-2}{y+1} = e^{3x + C}

\dfrac{y-2}{y+1} = Ce^{3x}

With the initial condition y(0) = 1, we find

\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12

so that the particular solution is

\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}

It's not too hard to solve explicitly for y; notice that

\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}

Then

1 - \dfrac3{y+1} = -\dfrac12 e^{3x}

\dfrac3{y+1} = 1 + \dfrac12 e^{3x}

\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}

y+1 = \dfrac6{2+e^{3x}}

y = \dfrac6{2+e^{3x}} - 1

\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}

7 0
2 years ago
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