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qaws [65]
3 years ago
14

Plzzzzzzzzz anyone helpppppp10points

Mathematics
1 answer:
BlackZzzverrR [31]3 years ago
5 0

when is 9 = 15.2 - 0.537x

         9 = 15.2 - 0.537x

  -15.2      -15.2                 subtract 15.2 from both sides

      -6.2   =   -0.537x  

/-0.537           /-0.537    divide -0.537 from both sides

11.55  = x   or 12 feet round to nearest whole foot


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Use the given graph to determine the limit, if it exists. A coordinate graph is shown with a horizontal line crossing the y axis
Lesechka [4]

Answer:

The limit of the function does not exists.

Step-by-step explanation:

From the graph it is noticed that the value of the function is 6 from all values of x which are less than 2. At x=2, the line y=6 has open circle. It means x=2 is not included.

For x<2

f(x)=6

The value of the function is -3 from all values of x which are greater than 2. At x=2, the line y=-3 has open circle. It means x=2 is not included.

For x>2

f(x)=-3

The value of y is 1 at x=2, because of he close circles on (2,1).

For x=2

f(x)=1

Therefore the graph represents a piecewise function, which is defined as

f(x)=\begin{cases}6& \text{ if } x2 \end{cases}

The limit of a function exist at a point a if the left hand limit and right hand limit are equal.

lim_{x\rightarrow a^-}f(x)=lim_{x\rightarrow a^+}f(x)

The function is broken at x=2, therefore we have to find the left and right hand limit at x=2.

lim_{x\rightarrow 2^-}f(x)=6

lim_{x\rightarrow 2^+}f(x)=-3

6\neq-3

Since the left hand limit and right hand limit are not equal therefore the limit of the function does not exists.

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Answer:

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Step-by-step explanation:

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Answer:

2x+6=-8(x>=-7), 2x+3<9(x<3), 2x-4<=-8(x<=-2), -x-8>=-5(x<=-3), x+-4<=-3(x<=1)

Step-by-step explanation:

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In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist
Tcecarenko [31]

Answer:

a. 0.7291

b. 0.9968

c. 0.7259

Step-by-step explanation:

a. np and n(1-p) can be calculated as:

np=23\times 0.48\\\\=11.04\\\\n(1-p)=23(1-0.52)\\\\=11.96

#Both np and np(1-p) are greater than 5, hence, normal approximation is most appropriate:

\mu_x=11.04\\\\\sigma^2=np(1-p)=0.48\times 0.52\times 23=5.7408

#Define Y:

Y~(11.04,5.7408)

P(X\leq 12)\approx P(Y\leq 12.5)\\\\P(Z\leq \frac{(12.5-11.04)}{\sqrt{5.7408}})=\\\\=1-0.2709\\\\=0.7291

Hence, the probability of 12 or fewer is 0.8291

b. The  probability that 5 or more fish were caught.

#Using normal approximation:

P(X \geq 5) \approx P(Y \geq 4.5) = P(Z \geq\frac{ (4.5-11.04)}{\sqrt{(5.7408)}})\\\\=1-0.0032\\\\=0.9968

Hence, the probability of catching 5+ is 0.9968

c. The probability of between 5 and 12 is calculated as;

-From b above P(X\geq 5)=0.9968 and a ,P(X\leq 12)=0.7291

P(5\leq X\leq 120\approx P(4.5\leq Y\leq  12.5)\\\\=0.7291-(1-0.9968)\\\\=0.7259

Hence, the probability of between 5 and 12 is 0.7259

4 0
3 years ago
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