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3 years ago

In David's bag of candy he has:

Mathematics

1 answer:

GaryK [48]3 years ago

4 0

Answer:

3/15 cause there are 15 candies in total and 3 snickers

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What is 278 divided by 2089

nekit [7.7K]

Answer:

0.13307802776

Step-by-step explanation:

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3 years ago

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Jake spent a three-day weekend on the lake. On Saturday, he rented a pontoon boat for three hours and paid $435. On Sunday, he r

Maksim231197 [3]

Answer:

Has nobody answered this?

Step-by-step explanation:

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3 years ago

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consider the distribution of monthly social security (OASDI) payments. Assume a normal distribution with a standard deviation of

Orlov [11]

Answer:

$\mu = x - z(\sigma)$

$\mu = 1214.87 - 0.67(116) \\\\\mu = 1214.87 - 77.72 \\\\\mu = \$1137.15$

Therefore, the mean monthly payment is $1137.15.

Step-by-step explanation:

What is Normal Distribution?

We are given a Normal Distribution, which is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.

We are asked to find the mean monthly social security (OASDI) payment.

Mean monthly payment = μ = ?

We are given that the standard deviation is $116

One-fourth of payments are above $1214.87

One-fourth means 25%

$P(X > x )= P(Z > z ) = 0.25\\\\P(X < x )= P(Z < z) = 1 - 0.25\\\\P(X < x )= P(Z < z) = 0.75\\\\$

From the z-table, the z-score corresponding to 0.75 is found to be 0.67

$z = 0.67$

The mean is found by

$x = \mu + z(\sigma)$

$\mu = x - z(\sigma)$

Where

x = $1214.87

z = 0.67

σ = $116

$\mu = 1214.87 - 0.67(116) \\\\\mu = 1214.87 - 77.72 \\\\\mu = \$1137.15$

Therefore, the mean monthly payment is $1137.15.

3 0

3 years ago

CAN SOMEONE HELP ME PLEASE ASAP!?

jekas [21]

Answer: 123 degrees

Step-by-step explanation:

hope this is the right answer :)

8 0

3 years ago

How can I evaluate this question?

ohaa [14]

$\bf \left(x^2-\cfrac{2}{\sqrt{x}}+1 \right)(\sqrt[3]{x}+3x-4)\quad 
\begin{cases}
\frac{2}{\sqrt{x}}\implies \frac{2}{x^{\frac{1}{2}}}\implies 2x^{-\frac{1}{2}}\\\\
\sqrt[3]{x}\implies x^{\frac{1}{3}}
\end{cases}
\\\\\\
(x^2-2x^{-\frac{1}{2}}+1)(x^{\frac{1}{3}}+3x-4)$

$\bf \\\\\\
\begin{cases}
x^2\cdot x^{\frac{1}{3}}+3x^3-4x^2\\\\
-2x^{-\frac{1}{2}}\cdot x^{\frac{1}{3}}-2x^{-\frac{1}{2}}\cdot 3x+2x^{-\frac{1}{2}}\cdot 4\\\\
+x^{\frac{1}{3}}+3x-4
\end{cases}
\\\\\\ 
\begin{cases}
x^{2+\frac{1}{3}}+3x^3-4x^2\\\\
-2x^{-\frac{1}{2}+\frac{1}{3}}-6x^{-\frac{1}{2}+1}+8x^{-\frac{1}{2}}\\\\
+x^{\frac{1}{3}}+3x-4
\end{cases}$

$\bf x^{\frac{7}{3}}+3x^3-4x^2-2x^{-\frac{1}{6}}-6x^{\frac{1}{2}}+8x^{-\frac{1}{2}}+x^{\frac{1}{3}}+3x-4
\\\\\\
\sqrt[3]{x^7}+3x^3-4x^2-\cfrac{2}{x^{\frac{1}{6}}}-6\sqrt{x}+\cfrac{8}{x^{\frac{1}{2}}}+\sqrt[3]{x}+3x-4
\\\\\\
x^2\sqrt[3]{x}+3x^3-4x^2-\cfrac{2}{\sqrt[6]{x}}-6\sqrt{x}+\cfrac{8}{\sqrt{x}}+\sqrt[3]{x}+3x-4$

8 0

3 years ago