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4 years ago

For which equation is (4, 3) a solution?

Mathematics

1 answer:

Slav-nsk [51]4 years ago

3 0

The answer is C because 2 x 4 is 8, minus 5 is 3.

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For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

C. $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

From the above expression, the power of (5) is 2

Express 2 as 6 - 4

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

Check the list of options for the expression on the left hand side

The correct answer is $(5-\frac{1}{2})^6$

3 0

3 years ago

Help me please I literally don’t understand any of this. I know it’s easy too.

Ber [7]

Holigina, this is the solution:

Let's recall that √x = x ^(1/2)

Therefore,

√x√x = x ^(1/2 + 1/2) = x^1 (We sum the exponents of the same base)

The correct answer is C. 1

5 0

1 year ago

If a well was 70-feet deep, the frog climbs 7 feet per hour, and it slips back 2 feet while resting? How long (in hours) will it

Amiraneli [1.4K]

Answer:

14 hours

Step-by-step explanation:

Given well is 70 ft deep.

Each step of the climb involves moving up at 7 ft/hr and then slips down 2 feet. Assuming the frog rests only once each hour, then the resultant progress made each hour,

= distance climbed - distance lost by slipping

= 7 - 2

= 5 ft / hr

Hence we can say that :

5 feet ---> takes 1 hour

1 foot ---> takes 1/5 hours

70 feet ---> takes (1/5) x 70= 14 hours

4 0

3 years ago

Read 2 more answers

What does a sneeze do? PLEASE HELP!!!!!!

AVprozaik [17]

Answer: The function of sneezing is to expel mucus containing irritants from the nasal cavity. A sneeze, or sternutation, is a semi-autonomous, convulsive expulsion of air from the lungs through the nose and mouth, usually caused by foreign particles irritating the nasal mucosa.

Step-by-step explanation:

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3 years ago

The fuel efficiency (in miles per gallon) of midsize minivans made by two auto companies is compared. Twenty test drivers are ra

Neporo4naja [7]

Answer:

$4.7-2.100*3.114*\sqrt{\frac{1}{10}+\frac{1}{10}}=1.775$

$4.7+2.100*3.114*\sqrt{\frac{1}{10}+\frac{1}{10}}=7.625$

So on this case the 95% confidence interval would be given by $1.775 \leq \mu_A -\mu_B \leq 7.625$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_A=57.9$ represent the sample mean A

$\bar X_B =53.2$ represent the sample mean B

n1=10 represent the sample A size

n2=10 represent the sample B size

$s_1 =3.4$ sample standard deviation for sample A

$s_2 =2.8$ sample standard deviation for sample B

$\mu_1 -\mu_2$ parameter of interest.

Solution to the problem

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

$(\bar X_A -\bar X_B) \pm t_{\alpha/2} s_p \sqrt{\frac{1}{n_A}+\frac{1}{n_B}}$ (1)

Where $s_p$ represent the standard deviation pooled given by:

$s_p =\sqrt{\frac{(n_A -1)s^2_A +(n_B -1)s^2_B}{n_A +n_B -2}}$

$s_p =\sqrt{\frac{(10 -1)(3.4)^2 +(10-1)(2.8)^2}{10 +10 -2}}=3.114$

The point of estimate for $\mu_A -\mu_B$ is just given by:

$\bar X_A -\bar X_B =57.9-53.2=4.7$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n_A +n_B -2=10+10-2=18$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,18)".And we see that $t_{\alpha/2}=2.10$

Confidence interval

Now we have everything in order to replace into formula (1):

$4.7-2.100*3.114*\sqrt{\frac{1}{10}+\frac{1}{10}}=1.775$

$4.7+2.100*3.114*\sqrt{\frac{1}{10}+\frac{1}{10}}=7.625$

So on this case the 95% confidence interval would be given by $1.775 \leq \mu_A -\mu_B \leq 7.625$

3 0

4 years ago