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3 years ago

7

a class of 54 students is divided into equal groups for orientation each group should have at least 7 students but no more than

10 students. What is the group size?

1 answer:

Dennis_Churaev [7]3 years ago

8 0

2 students more were introducced in the class and each group had 8 people

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What is the equation of the line that passes through the point (-6, -8) and has a slope of 1/3

ivanzaharov [21]

Answer <u>(assuming it can be in point-slope form)</u>:

$y + 8 = \frac{1}{3} (x+6)$

Step-by-step explanation:

With the given information, we can use the point-slope formula, $y-y_1 = m (x-x_1)$ , to write the equation of the line. Substitute values for the $m$ , $x_1$ , and $y_1$ in the formula to do so.

The $m$ represents the slope, so substitute $\frac{1}{3}$ in its place. The $x_1$ and $y_1$ represent the x and y values of one point the line intersects, so substitute -6 for $x_1$ and -8 for $y_1$ . This gives the following answer and equation (just make sure to convert the double negatives into positives:

$y-(-8) = \frac{1}{3} (x-(-6))\\y + 8 = \frac{1}{3} (x+6)$

3 0

3 years ago

Read 2 more answers

Enter the sum of numbers as a product of their GCF. 75 + 30 The sum of the numbers as a product of their GCF is

Oksanka [162]

Answer: 75+30 = 15 x 7

Step-by-step explanation:

The given expression is 75+30 (=105) which defines the sum of 75 and 30.

Prime factorization of 75 and 30 are as below:

75 = 5 x 5 x 3

30 = 5 x 3 x 2

GCD (75,30) = 5x 3 = 15 [Note: GCD = Greatest common divisor]

Consider 75+30 = (15 x 5) + (15 x 2) [75 = 15 x 5 and 30= 15 x 2]

= 15 (5+2) [taking 15 as common ]

= 15 x (7)

(=105)

So, 75+30 which is sum of the numbers and it is expressed as 15 x 7 which a product of their GCF.

7 0

3 years ago

Suppose we roll a fair die and let X represent the number on the die. (a) Find the moment generating function of X. (b) Use the

Likurg_2 [28]

Answer:

(a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{2 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

Step-by step explanation:

Given X represents the number on die.

The possible outcomes of X are 1, 2, 3, 4, 5, 6.

For a fair die, $P(X)=\frac{1}{6}$

(a) Moment generating function can be written as $M_{x}(t)$ .

$M_x(t)=\sum_{x=1}^{6} P(X=x)$

$M_{x}(t)=\frac{1}{6} e^{t}+\frac{1}{6} e^{2 t}+\frac{1}{6} e^{3 t}+\frac{1}{6} e^{4 t}+\frac{1}{6} e^{5 t}+\frac{1}{6} e^{6 t}$

$M_x(t)=\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) Now, find $E(X) \text { and } E\((X^{2})$ using moment generating function

$M^{\prime}(t)=\frac{1}{6}\left(e^{t}+2 e^{2 t}+3 e^{3 t}+4 e^{4 t}+5 e^{5 t}+6 e^{6 t}\right)$

$M^{\prime}(0)=E(X)=\frac{1}{6}(1+2+3+4+5+6)$

$\Rightarrow E(X)=\frac{21}{6}$

$M^{\prime \prime}(t)=\frac{1}{6}\left(e^{t}+4 e^{2 t}+9 e^{3 t}+16 e^{4 t}+25 e^{5 t}+36 e^{6 t}\right)$

$M^{\prime \prime}(0)=E(X)=\frac{1}{6}(1+4+9+16+25+36)$

$\Rightarrow E\left(X^{2}\right)=\frac{91}{6}$

Hence, (a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$ .

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

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4 years ago

Assume the cost of a car is $25,000. With continuous compounding, in effect, the cost of the care will increase according to the

Masja [62]

The value of t that makes the factor e^(.032t) have the value of 2 can be found using logarithms.
2 = e^(0.032t)
ln(2) = ln(e^(0.032t)) = 0.032t
t = ln(2)/0.032 ≈ 21.66

It would take 21.66 years for the cost to double.

4 0

3 years ago

Find the area.<br> 5.cm<br> 5 cm<br> 3 cm<br> 8 cm

Talja [164]

Answer: 600

Step-by-step explanation:

5x5=25

3x8=24

25x24=600

7 0

3 years ago

Read 2 more answers