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3 years ago

A. 12.4 b. 12.9 c. 8.5 d. 8.1

Mathematics

2 answers:

12345 [234]3 years ago

8 0

It would be c because it djfjfjd. F fjdjfjf

Amiraneli [1.4K]3 years ago

6 0

I say 8.5 not sure so c.

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culture started with 3,000 bacteria. After 2 hours it grew to 3,300 bacteria. Predict how many bacteria will be present after 16

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A Function is a rule that assigns each value of the independent variable to exactly _____ value of the dependent variable

kherson [118]

A function is a rule that assigns each value of the independent variable to exactly ONE value of the dependant variable

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4 years ago

Sound City sells the ClearTone-400 satellite car radio. For this radio, historical sales records over the last 100 weeks show 4

Ksivusya [100]

Answer:

See below

Step-by-step explanation:

$\bf \mu(x)$ is the average

$\bf \mu(x)=\frac{4*0+33*1+39*2+12*3+8*4+4*5}{100}=\frac{199}{100}=1.99\;radios/week$

$\bf \sigma^2(x)$ is the variance

$\bf \sigma^2(x)=\frac{4*(0-1.99)^2+33*(1-1.99)^2+39*(2-1.99)^2+12*(3-1.99)^2+8*(4-1.99)^2+4*(5-1.99)^2}{100}=\frac{128.99}{100}=1.2899$

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4 years ago

Which equation represents a hyperbola with a center at (0, 0), a vertex at (−48, 0), and a focus at (50, 0)?

Lerok [7]

bearing in mind that "a" is the length of the traverse axis, and "c" is the distance from the center to either foci.

we know the center is at (0,0), we know there's a vertex at (-48,0), from the origin to -48, that's 48 units flat, meaning, the hyperbola is a horizontal one running over the x-axis whose a = 48.

we also know there's a focus point at (50,0), that's 50 units from the center, namely c = 50.

$\bf \textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2}\\ \textit{asymptotes}\quad y= k\pm \cfrac{b}{a}(x- h) \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \begin{cases} h=0\\ k=0\\ a=48\\ c=50 \end{cases}\implies \cfrac{(x-0)^2}{48^2}-\cfrac{(y-0)^2}{b^2}=1 \\\\\\ c=\sqrt{a^2+b^2}\implies \sqrt{c^2-a^2}=b\implies \sqrt{50^2-48^2}=b \\\\\\ \sqrt{196}=b\implies 14=b~\hspace{3.5em}\cfrac{(x-0)^2}{48^2}-\cfrac{(y-0)^2}{14^2}=1\implies \cfrac{x^2}{48^2}-\cfrac{y^2}{14^2}=1$

8 0

3 years ago

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