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2 years ago

Find the value of each expression. Show your work. 10^C^4

1 answer:

6 0

$_nC_r = \frac{n!}{(n - r)!r!}$
<span> $_1_0C_4 = \frac{10!}{(10 - 4)!4!}$
</span> $_1_0C_4 = \frac{10!}{6!4!}$
$_1_0C_4 = \frac{10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1}{6 * 5 * 4 * 3 * 2 * 1 * 4 * 3 * 2 * 1}$
$_1_0C_4 = \frac{10 * 9 * 8 * 7}{4 * 3 * 2 * 1}$
$_1_0C_4 = \frac{90 * 56}{12 * 2}$
$_1_0C_4 = \frac{5040}{24}$
$_1_0C_4 = 210$

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Step-by-step explanation:

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2 years ago

Read 2 more answers

A simple random sample was taken of 44 water bottles from a bottling plant’s warehouse. The dissolved oxygen content (in mg/L) w

Margarita [4]

Answer:

(a) Test statistic is -2.85 and p-value is 0.0022

(b) Reject the null hypothesis. The population mean of dissolved oxygen content is not equal to 10 mg/L

Step-by-step explanation:

H0: mu equals 10

Ha: mu not equals 10

The test is a two-tailed test because the alternate hypothesis is expressed using not equal to

(a) Test statistic (z) = (sample mean - population mean) ÷ (sd/√n) = (9.14 - 10) ÷ (2/√44) = -0.86 ÷ 0.302 = -2.85

Cumulative area of the test statistic = 0.9978

p-value = 2(1 - 0.9978) = 2(0.0022) = 0.0044

(b) The critical value using 0.02 significance level is 2.422. For a two-tailed test, the region of no rejection of the test statistic lies between -2.422 and 2.422.

Conclusion:

Reject the null hypothesis because the test statistic -2.85 falls outside the region bounded by the critical values -2.422 and 2.422.

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2 years ago

There are 4 red marbles and 8 green marbles in a box. If Emma randomly selects a marble from the box, what is the probability of

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