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3 years ago

A square has a perimeter of 14.4 in. What is the length of each of the sides?

Mathematics

2 answers:

Tju [1.3M]3 years ago

7 0

The length of each side is 3.6

scZoUnD [109]3 years ago

3 0

14.4/4 is 3.6. Therefore the answer is 3.6.

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Gabrielle took a Spanish test with 50 questions and answered 44 of them correctly. What percent of the questions did Gabrielle a

mrs_skeptik [129]

She answered 88% correct. 44 times 2 is 88.

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3 years ago

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What is the prime factorization of 200? 2 2 × 5 3 2 3 × 5 2 2 4 × 5 2 × 5 4

Misha Larkins [42]

Answer:

2^3 * 5^2

Step-by-step explanation:

We need to break 200 down into prime numbers

200 = 20*10 These are not prime

= 4*5 * 2*5 The 4 is not prime

= 2*2*5 * 2*5 This is prime

Rearranging

= 2*2*2*5*5

= 2^3 * 5^2

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3 years ago

The average score of this quiz is 24 points with a standard deviation of 1.5 points. What is the probability that someone scored

navik [9.2K]

Answer:

8.08 %

Step-by-step explanation:

Let us denote : $X=$ score X follows a normal distribution with mean $=24$ and standard deviation $=1.5$

Step 2

The probability that someone scored 21.9 or less is :

$\mathrm{P}(\mathrm{X} \leq 21.9)$

$=\mathrm{P}\left(\mathrm{Z} \leq \frac{21.9-24}{1.5}\right)$

$=\mathrm{P}(\mathrm{Z} \leq-1.4)$

$=0.0808$

$=8.08%$

Answer : $8.08 %$

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3 years ago

(b) dy/dx = (x - y+ 1)^2

Elanso [62]

Substitute $v(x)=x-y(x)+1$ , so that

$\dfrac{\mathrm dv}{\mathrm dx}=1-\dfrac{\mathrm dy}{\mathrm dx}$

Then the resulting ODE in $v(x)$ is separable, with

$1-\dfrac{\mathrm dv}{\mathrm dx}=v^2\implies\dfrac{\mathrm dv}{1-v^2}=\mathrm dx$

On the left, we can split into partial fractions:

$\dfrac12\left(\dfrac1{1-v}+\dfrac1{1+v}\right)\mathrm dv=\mathrm dx$

Integrating both sides gives

$\dfrac{\ln|1-v|+\ln|1+v|}2=x+C$

$\dfrac12\ln|1-v^2|=x+C$

$1-v^2=e^{2x+C}$

$v=\pm\sqrt{1-Ce^{2x}}$

Now solve for $y(x)$ :

$x-y+1=\pm\sqrt{1-Ce^{2x}}$

$\boxed{y=x+1\pm\sqrt{1-Ce^{2x}}}$

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3 years ago

A wrestling match has eliminations each round. The table below shows the number of wrestlers in each of the first 4 rounds of th

AlladinOne [14]

$\bf \begin{array}{lllll}
round(x)&\boxed{1}&2&3&\boxed{4}\\\\
wrestlers[f(x)]&\boxed{64}&32&18&\boxed{9}
\end{array}
\\\\\\
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------\\\\
f(x)= \qquad 
\begin{cases}
x_1=1\\
x_2=4
\end{cases}\implies \cfrac{f(4)-f(1)}{4-1}\implies \cfrac{9-64}{4-1}\implies \cfrac{-55}{3}$

55 over 3, or 55 wrestlers for every 3 rounds, but the wrestlers value is negative, thus 55 "less" wrestlers for every 3 rounds on average.

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3 years ago

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