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3 years ago

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Evaluate 7c4 a. 6 b. 1 c. 35 d. 840

2 answers:

valentinak56 [21]3 years ago

6 0

Hello,

7C4=7*6*5*4/(4*3*2*1)=35

Answer C

Yuki888 [10]3 years ago

3 0

Answer:

$^7C_4 = 35$

Step-by-step explanation:

We are given the following information in the question:

We have to evaluate:

Formula:

$^7C_4\\^n C_x = \displaystyle\frac{n!}{(n-x)!x!}\\\\^7C_4 = \displaystyle\frac{7!}{(7-4)!4!} =\frac{7!}{3!4!} =\frac{7.6.5.4.3.2.1}{3.2.1.4.3.2.1} = 35$

Thus, the is values is 35

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Which of the following phrases translates to m - 7?

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Answer: Both phrase A and phrase D translate to m - 7

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Stewart is making banana pudding for a family reunion. Her recipe uses 1.5 bananas per dish of pudding. If one dish of pudding c

Rina8888 [55]

Answer: She will need about 15 bananas to feed 250 people that will be at the reunion.

Step-by-step explanation:

Let be "x" the number of bananas Stewart will need to feed 250 people that will be at the reunion.

According to the infomation provided in the exercise, we know that she needs 1.5 bananas per dish of pudding and one dish of pudding can feed 25 people.

Knowing this, we can set up the following proportion:

$\frac{1.5}{25}=\frac{x}{250}$

Finally, we must solve for "x":

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3 years ago

A new post-surgical treatment is being compared with a standard treatment. Seven subjects receive the new treatment, while seven

a_sh-v [17]

Answer:

$(17.714-28.714) -2.681 \sqrt{\frac{4.461^2}{7} +\frac{7.387^2}{7}}= -19.745$

$(17.714-28.714) +2.681 \sqrt{\frac{4.461^2}{7} +\frac{7.387^2}{7}}= -2.255$

Step-by-step explanation:

For this case we have the following info given:

Treatment: 12 13 15 19 20 21 24

Control: 18 23 24 30 32 35 39

We can find the sample mean and deviations with the the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s =\sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}$

And repaplacing we got:

$\bar X_T = 17.714$ the sample mean for treatment

$\bar X_C = 28.714$ the sample mean for treatment

$s_T= 4.461$ the sample deviation for treatment

$s_C= 7.387$ the sample deviation for control

$n_T= n_C= 7$ the sample size for each sample

The degrees of freedom are given by:

$df= 7+7-2= 12$

The confidence interval for the difference of means is given by:

$(\bar X_T -\bar X_C) \pm t_{\alpha/2} \sqrt{\frac{s^2_T}{n_T} +\frac{s^2_C}{n_C}}$

The confidence is 98% so then the significance is $\alpha=0.02$ and $\alpha/2 =0.01$ . Then the critical value would be:

$t_{\alpha/2}=2.681$

And replacing we got:

$(17.714-28.714) -2.681 \sqrt{\frac{4.461^2}{7} +\frac{7.387^2}{7}}= -19.745$

$(17.714-28.714) +2.681 \sqrt{\frac{4.461^2}{7} +\frac{7.387^2}{7}}= -2.255$

4 0

3 years ago

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joja [24]

Answer:

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Step-by-step explanation:

Greg = g

Teri = t

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Formulas:

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Answer:

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Check:

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4 years ago

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