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2 years ago

10

A motion picture camera at normal speed takes 24 pictures per second. How many pictures are in a movie that is 90 minutes long?

(Hint 1min = 60 seconds)

2 answers:

ad-work [718]2 years ago

4 0

24 × 60 = 1,440
1,440 × 90 = 129,600

There are 129,600 pictures in a movie that is 90 minutes long.

TiliK225 [7]2 years ago

3 0

P = 24 (90 x 60)
P = 24 x 5400
P = 129600

129600 pictures are taken during a 90 minute movie.

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A spinner has 16 equal-sized sections. Twelve of the sections are orange.

Usimov [2.4K]

The probability that the spinner will land on orange is the likelihhood of landing on orange

The probability that the spinner will land on orange is 0.75

<h3>How to determine the probability?</h3>

From the question, we have:

Spinner section = 16

Orange section = 12

So, the probability that the spinner will land on orange is:

p = 12/16

Evaluate

p = 0.75

Hence, the probability that the spinner will land on orange is 0.75

Read more about probability at:

brainly.com/question/251701

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2 years ago

the sum of 3 consecutive integers is 29 less than 5 times the smallest number. find the three numbers

allochka39001 [22]

x + x + 1 + x + 2 = 5x - 29

Combine like terms.

3x + 3 = 5x - 29

Subtract 3x from both sides.

3 = 2x - 29

Add 29 to both sides.

32 = 2x

Divide both sides by 2.

16 = x

<h3>The first digit is 16.</h3><h3>The second digit is 17.</h3><h3>The third digit is 18.</h3>

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3 years ago

Solve the following equations below by showing all the steps.<br> - 1. 9(4 - 2x) – 3 = 4 - 6(3x – 5)

In-s [12.5K]

Answer:

No real solution

Step-by-step explanation:

First off I'm assuming that the 1 or -1 at the begging is just the problem number:

First you distribute on both sides

36 - 18x - 3 = 4 - 18x + 30

Then combine like terms

33 - 18x = 34 - 18x

Then rearrange the terms by adding 18x to both sides and subtracting both sides by 33. You should get:

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3 years ago

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Anna [14]

Answer:

a) 0.4452

b) 0.0548

c) 0.0501

d) 0.9145

e) 6.08 minutes or greater

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.7 minutes

Standard Deviation, σ = 0.50 minutes.

We are given that the distribution of length of the calls is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(calls last between 4.7 and 5.5 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{4.7 - 4.7}{0.50} \leq z \leq \displaystyle\frac{5.5-4.7}{0.50}) = P(0 \leq z \leq 1.6)\\\\= P(z \leq 1.6) - P(z$

$P(4.7 \leq x \leq 5.5) = 44.52\%$

b) P(calls last more than 5.5 minutes)

$P(x > 5.5) = P(z > \displaystyle\frac{5.5-4.7}{0.50}) = P(z > 1.6)\\\\P( z > 1.6) = 1 - P(z \leq 1.6)$

Calculating the value from the standard normal table we have,

$1 - 0.9452 = 0.0548 = 5.48\%\\P( x > 5.5) = 5.48\%$

c) P( calls last between 5.5 and 6 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{5.5 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(1.6 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$

$P(5.5 \leq x \leq 6) = 5.01\%$

d) P( calls last between 4 and 6 minutes)

$P(4 \leq x \leq 6) = P(\displaystyle\frac{4 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(-1.4 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$

$P(4 \leq x \leq 6) = 91.45\%$

e) We have to find the value of x such that the probability is 0.03.

P(X > x)

$P( X > x) = P( z > \displaystyle\frac{x - 4.7}{0.50})=0.03$

$= 1 -P( z \leq \displaystyle\frac{x - 4.7}{0.50})=0.03$

$=P( z \leq \displaystyle\frac{x - 4.7}{0.50})=0.997$

Calculation the value from standard normal z table, we have,

P(z < 2.75) = 0.997

$\displaystyle\frac{x - 4.7}{0.50} = 2.75\\x = 6.075 \approx 6.08$

Hence, the call lengths must be 6.08 minutes or greater for them to lie in the highest 3%.

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2 years ago

Describe the translation of the function from the parent function f(x) = x².

kkurt [141]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\

\end{array}\\$

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\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}
$

now, notice the template above... now let's see your function $\bf y=x^2+4\implies 
\begin{array}{llllll}
y=&1(&1x&+&0)^2+&4\\
&A&B&&C&D
\end{array}$

so.. what do you think was the shift then?

8 0

2 years ago