3 years ago

What is (5 radical 2) squared?

Mathematics

1 answer:

Roman55 [17]3 years ago

3 0

(5 radical 2)^2 =

5^2 × (radical 2)^2 =

25 × 2 = 50

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A certain university has 10 vehicles available for use by faculty and staff. four of these are vans and 6 are cars. on a particu

REY [17]

There are 4 vans. So we have that probability that the first vehicle is a van p (e) = 4/10 = 0.4. P(e|f) = P( f and e) / p (f) p(e) = 0.4 and p(f) = 3/9 P (f and e) = 0.40 * 0.33 = 0.132 So p(e|f) = 0.4 * 0.33/ 0.33 = 0.4 P(f and e) p(f) * p(e) = 0.4 * 0.33 = 0.132

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3 years ago

Solve for y 5y-x=10 Is it 2?

svet-max [94.6K]

5y-x=10
Isolate the variable
add x to both sides
5y-x+x=10+x
5y=10+x
Divide both sides by 5
5y/5=10+x/5
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So y is equal to 2+1/5x

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2 years ago

Solve the system x+5y=-15 -2x-10y=30

skelet666 [1.2K]

X= -2x -15y -45 maybe i’m kinda stupid

8 0

2 years ago

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A veterinarian collects data on the number of times race horses are raced during their careers. The veterinarian finds that the

SpyIntel [72]

Answer:

c. (12.12, 18.48)

Step-by-step explanation:

Hello!

The study variable is X: number of times a racehorse is raced during its career.

The average number is X[bar]= 15.3 and the standard deviation is S= 6.8 obtained from a sample of n=20 horses.

To estimate the population mean you need that the variable has a normal distribution, in this case, we have no information about its distribution so I'll assume that it has a normal distribution. With n=20 the most accurate statistic to use for the estimation is a Students-t for one sample, the formula for the interval is:

X[bar] ± $t_{n-1;1-\alpha /2} * \frac{S}{\sqrt{20} }$

$t_{n-1;1-\alpha /2} = t_{19;0.975}= 2.093$

[15.3 ± 2.093 * $\frac{6.8}{\sqrt{20} }$ ]

[12.12; 18.48]

Using a significance level of 95% you'd expect that the true average of times racehorses are raced during their career is included in the interval [12.12; 18.48].

I hope it helps!

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3 years ago

Need help with this question. Thank you

V125BC [204]

I think the answer is b

3 0

3 years ago

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