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3 years ago

What is -5/4 to the 2nd power?

Mathematics

1 answer:

Vsevolod [243]3 years ago

8 0

Answer:

$\frac{25}{16}$

Step-by-step explanation:

$(-\frac{5}{4})^2\\\\ \text {Apply power of a fraction rule: } (\frac{a}{b})^x=\frac{a^x}{b^x}\\\\(-\frac{5}{4})^2 = \frac{-5^2}{4^2}=\frac{25}{16}\\\\\boxed{(-\frac{5}{4})^2=\frac{25}{16}}$

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What number increased by 7 equals two times the number? 6 7 8 9 11

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Answer:

Step-by-step explanation:

7 plus 7 equals 14, which is two times 7

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2 years ago

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For the graffiti cat sweater on page 9, Dodd knit

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Answer: 54 stitches

Step-by-step explanation:

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Help!! BRAINLIEST TOO!

natulia [17]

Answer:

36.5 inches

Step-by-step explanation:

Given

See attachment for the given data

Required

Which length is closest to 4.2lb

The given data is a linear dataset.

So, we start by calculating the slope (m)

$m =\frac{y_2 - y_1}{x_2 - x_1}$

Pick any two corresponding points from the table

So, we have:

$(x_1,y_1) = (8,1.21)$

$(x_2,y_2) = (12,1.63)$

So:

$m =\frac{y_2 - y_1}{x_2 - x_1}$

$m =\frac{1.63 - 1.21}{12 - 8}$

$m =\frac{0.42}{4}$

$m =0.105$

The linear equation is then calculated using:

$y = m(x - x_1) + y_1$

This gives:

$y = 0.105 * (x - 8) + 1.21$

Open bracket

$y = 0.105x - 0.84 + 1.21$

$y = 0.105x + 0.37$

To get the length closest to 4.2lb,

we set $y = 4.2$

Then solve for x

So, we have:

$y = 0.105x + 0.37$

$4.2 = 0.105x + 0.37$

Collect like terms

$0.105x= 4.2 - 0.37$

$0.105x= 3.83$

Solve for x

$x= 3.83/0.105$

$x\approx 36.5$

4 0

3 years ago

Given function of f(x) = 2x - 1 and fg(x) = 10x + 3, find the function of g(x)

denis-greek [22]

Answer:

The function of g(x) = 5x + 2

Step-by-step explanation:

Let us use the composite function to solve the question

∵ f(x) = 2x - 1

∵ f(g(x)) = 10x + 3

→ f(g(x)) means substitute x in f(x) by g(x)

∴ f(g(x)) = 2[g(x)] - 1

→ Equate the two right sides of f(g(x))

∴ 2[g(x)] - 1 = 10x + 3

→ Add 1 to both sides

∴ 2[g(x)] - 1 + 1 = 10x + 3 + 1

∴ 2[g(x)] = 10x + 4

→ Divide each term into both sides by 2

∵ $\frac{2[g(x)]}{2}$ = $\frac{10x}{2}$ + $\frac{4}{2}$

∴ g(x) = 5x + 2

∴ The function of g(x) = 5x + 2

7 0

2 years ago

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

Elenna [48]

Answer:

Part a: The probability of no arrivals in a one-minute period is 0.000045.

Part b: The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c: The probability of no arrivals in a 15-second is 0.0821.

Part d: The probability of at least one arrival in a 15-second period is 0.9179.

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

$X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)$

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

Substitute the value of λ=10 in the formula as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}$

Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}$

The probability of no arrivals in a one-minute period is 0.000045.

Part b:

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

$\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}$

The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as $\frac{X}{4}$

$\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}$

That is,

$Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)$

So, the probability mass function of Y is,

$P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots$

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}$

The probability of no arrivals in a 15-second is 0.0821.

Part d:

The probability that there is at least one arrival in a 15-second period is calculated as,

$\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}$

$\begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}$

The probability of at least one arrival in a 15-second period is 0.9179.

7 0

3 years ago