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Goshia [24]
3 years ago
7

Number of the day 3.59

Mathematics
1 answer:
sleet_krkn [62]3 years ago
7 0
Please be a little specific.
You might be interested in
One more time!
CaHeK987 [17]
Since q(x) is inside p(x), find the x-value that results in q(x) = 1/4

\frac{1}{4} = 5 - x^2\ \Rightarrow\ x^2 = 5 - \frac{1}{4}\ \Rightarrow\ x^2 = \frac{19}{4}\ \Rightarrow \\
x = \frac{\sqrt{19} }{2}

so we conclude that
q(\frac{\sqrt{19} }{2} ) = 1/4

therefore

p(1/4) = p\left( q\left(\frac{ \sqrt{19} }{2} \right)  \right)

plug x=\sqrt{19}/2 into p( q(x) ) to get answer

p(1/4) = p\left( q\left( \frac{ \sqrt{19} }{2} \right) \right)\ \Rightarrow\ \dfrac{4 - \left(  \frac{\sqrt{19} }{2}\right)^2 }{ \left(  \frac{\sqrt{19} }{2}\right)^3 } \Rightarrow \\ \\ \dfrac{4 - \frac{19}{4} }{ \frac{19\sqrt{19} }{8}} \Rightarrow \dfrac{8\left(4 - \frac{19}{4}\right) }{ 8 \cdot \frac{19\sqrt{19} }{8}} \Rightarrow \dfrac{32 - 38}{19\sqrt{19}} \Rightarrow \dfrac{-6}{19\sqrt{19}} \cdot \frac{\sqrt{19}}{\sqrt{19}}\Rightarrow

\dfrac{-6\sqrt{19} }{19 \cdot 19} \\ \\ \Rightarrow  -\dfrac{6\sqrt{19} }{361}

p(1/4) = -\dfrac{6\sqrt{19} }{361}
3 0
3 years ago
Responda abaixo: a)2x+3y=10*√4 b)x+2y+5*3=x+y
RoseWind [281]

Answer:

a) x = 10 + -1.5y

b) y+15

Step-by-step explanation:

a)2x+3y=10*√4

2x+3y=20

2x + 3y + -3y = 20 + -3y

2x + 0 = 20 + -3y

2x = 20 + -3y

x ÷ 2 = (20 + -3y) ÷2

x = 10 + -1.5y

b)x+2y+5*3=x+y

x+2y+15=x+y

x+2y+15 - y=x+y - y

x+y+15 =x

x+y+15 - x  =x - x

y+15

5 0
4 years ago
4 ten thousands 3 ten thousands 4 hundred 8 tens
iVinArrow [24]
That would be 3'000 + 4000 + 400 + 80 = 7,480
6 0
3 years ago
Log2 (M) + log3 (N)<br> how do you solve??
natta225 [31]

Answer:

log2(M)+log3(N)

simplifies to:

simplified step by step

Step-by-step explanation:

0.30103m+0.477121n.

the answer is

0.39103m+0.477121n

6 0
3 years ago
Suppose the FAA weighed a random sample of 20 airline passengers during the summer and found their weights to have a sample mean
Orlov [11]

Answer:

Step-by-step explanation:

Given Parameters

Mean, x = 180

total samples, n = 20

Standard dev, \sigma = 30

\alpha = 1 - 0.95 = 0.05 at 95% confidence level

Df = n - 1 = 20 - 1 = 19

Critical Value, t_\alpha, is given by

t_{c}=t_{\alpha, df} = t_{0.05,19} = 1.729

a).

Confidence Interval, \mu, is given by the formula

\mu = x +/- t_c \times \frac{s}{\sqrt{n} }

\mu = 180 +/- 1.729 \times \frac{30}{\sqrt{20} }

\mu = 180 +/-11.5985

191.5985 > \mu > 168.4015

b).

Critical Value, t_{\alpha/2}, is given by

t_{c}=t_{\alpha/2, df} = t_{0.05/2,19} = 2.093

Confidence Interval, \mu, is given by

\mu = x +/- t_c \times \frac{s}{\sqrt{n} }

\mu = 180 +/- 2.093 \times \frac{30}{\sqrt{20} }

    = 180 +/- 14.0403

    = 165.9597 < \mu < 194.0403

8 0
3 years ago
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