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3 years ago

Find the derivative of the function y = / - 5 - 3x. dy/dx =

Mathematics

2 answers:

jeyben [28]3 years ago

6 0

-3 just here to reassure you

Katena32 [7]3 years ago

3 0

Answer:

-3

Step-by-step explanation:

use the power rule to solve this

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Ms. Carol sold 346 child tickets and 253 adult tickets. How many tickets did Ms. Carol sell?

pickupchik [31]

She sold 346+253=599 tickets.

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3 years ago

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Lunna [17]

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Answer:

(c) The volume of a rectangular pyramid is 1/3 the volume of a rectangular prism. The volume of the rectangular pyramid is 66 cm3.

Step-by-step explanation:

The pyramid with the same overall dimensions will obviously have a smaller volume. The ratio is 1/3.

The formula for the volume of a rectangular prism is ...

V = LWH

The formula for the volume of a rectangular pyramid is ...

V = (1/3)LWH

The pyramid has 1/3 the volume of the prism.

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3 years ago

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What is the answer to the blank 1 3 5 2 4 _ (It is not 6)

Makovka662 [10]

It might be 9 if it's not 6 ...but I can't see how it wouldn't be 6?

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3 years ago

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What is the answer to this equation <img src="https://tex.z-dn.net/?f=10%5Csqrt%7Bx%7D%2020" id="TexFormula1" title="10\sqrt{x}

AveGali [126]

Factor the expression
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3 years ago

A four year old is going to spin around with his arms stretched out 100 times. From past experience, his father knows it takes a

Sati [7]

Answer:

$P(X \geq0.55) \leq 0.22$

Step-by-step explanation:

Using central Limit Theorem (CLT), The sum of 100 random variables;

$Y=X_1+X_2+...+X_{100}$ is approximately normally distributed with

Y ~ N (100 × $\frac{1}{3^2}$ ) = N ( 50, $\frac{100}{9}$ )

The approximate probability that it will take this child over 55 seconds to complete spinning can be determined as follows;

N ( 50, $\frac{100}{9}$ )

$P(Y>55) =P(Z>\frac{55-50}{10/3})$

$P(Y>55) =P(Z>1.5)$

$P(Y>55) =\phi (-1.5)$

$P(Y>55) =0.0668$

Using Chebyshev's inequality:

$P(|X-\mu\geq K)\leq \frac{\sigma^2}{K^2}$

Let assume that X has a symmetric distribution:

Then:

$2P(X-\mu\geq K)\leq) \frac{\sigma^2}{K^2}$

$2P(X \geq \mu+K)\leq) \frac{\sigma^2}{K^2}$

$2P(X\geq0.5+0.05)\leq \frac{\frac{1}{\frac{3^2}{100} } }{0.05^2}$ where: ( $\sigma^2 = \frac{1}{3^2/100}$ )

$P(X \geq0.55) \leq 0.22$

6 0

3 years ago