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3 years ago

Evaluate 19C1 PLEASE HELP

Mathematics

2 answers:

puteri [66]3 years ago

7 0

Answer:

Your answer will be 19C1 =19

LiRa [457]3 years ago

5 0

Answer:

${19}C_1=19$

Step-by-step explanation:

We need to find the value of ${19}C_1$ .

C stands for combination.

The formula of combination is as follows :

$nC_r=\dfrac{n!}{r!(n-r)!}$

Here,

n = 19 and r = 1

So,

$nC_r=\dfrac{19!}{1!(19-1)!}\\\\nC_r=\dfrac{19!}{1!\times 18!}\\\\nC_r=\dfrac{19\times 18!}{1!\times 18!}\\\\nC_r=19$

So, the value of ${19}C_1$ is 19.

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DochEvi [55]

Answer:

$C\approx 5.14$

Step-by-step explanation:

The circumference is the distance around a complete circle: in essence, the perimeter of a circle. The formula to find the circumference of a circle is the following:

$C=(d)(\pi)$

Where (d) represents the diameter, the largest cord that can be drawn in a circle, it goes from one end to another, passing through the center of the circle. The parameter ( $\pi$ ) represents the number (3.1415...). However, one is given a semicircle, thus one must divide the result in two, and then add the value of the diameter to find the perimeter of the given object. Therefore the equation for the circumference of this circle is as follows:

$C=\frac{(d)(\pi)}{2}+(d)$

Substitute in the given value,

$C=\frac{(2)(3.1415)}{2}+(2)$

Simplify,

$C=3.1415...+2$

$C=5.1415$

$C\approx 5.14$

3 0

3 years ago

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

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Hoochie [10]

Answer:

sq rt of 66 is 8.12 so 8.1 closest

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2 years ago

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Step-by-step explanation:

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3 years ago