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1 year ago

Form a sequence that has two arithmetic means between -1 and 59.

2 answers:

7 0

The formula for solving the problem is as follow:
an = a1 + (n - 1)d
Where:
n = number of figure in the sequence = 4
d = difference between successive number =?
a1 = -1
a4 = 59
Insert the given values into the formula,
59 = -1 + (4 - 1)d
59 = -1 + 3d
59 + 1 = 3d
60 = 3d
d = 60/3 = 20
Therefore, d = 20. This implies that, there is a difference of 20 between successive numbers.
The number sequence is as follow:
-1, 19, 39, 59.

Stels [109]1 year ago

7 0

two arithmetic means between -1 and 59.
b.-1, 19, 39, 59

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Find a line that lies entirely in the set defined by the equation x2 + y2 − z2 = 1.

RUDIKE [14]

Answer:

$x=1\,,\,y=t\,,\,z=t$

Step-by-step explanation:

A line is a one-dimensional figure that has no thickness and extends infinitely in both directions.

A line segment is a line that has two end-points and a ray is a line that has one end-point.

Given equation is $x^2+y^2-z^2=1$

To find : a line that lies entirely in the set defined by the given equation.

Solution:

Take $x=1\,,\,y=t\,,\,z=t$

Check:

$L.H.S=1^2+t^2-t^2=1+0=1=R.H.S$

Therefore, $x=1\,,\,y=t\,,\,z=t$ satisfy the given equation.

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1 year ago

(Ross 5.15) If X is a normal random variable with parameters µ " 10 and σ 2 " 36, compute (a) PpX ą 5q (b) Pp4 ă X ă 16q (c) PpX

pochemuha

Answer:

(a) 0.7967

(b) 0.6826

(d) 0.9525

(e) 0.1587

Step-by-step explanation:

The random variable X follows a Normal distribution with mean μ = 10 and variance σ² = 36.

(a)

Compute the value of P (X > 5) as follows:

$P(X>5)=P(\frac{x-\mu}{\sigma}>\frac{5-10}{\sqrt{36}})\\=P(Z>-0.833)\\=P(Z$

Thus, the value of P (X > 5) is 0.7967.

(b)

Compute the value of P (4 < X < 16) as follows:

$P(4$

Thus, the value of P (4 < X < 16) is 0.6826.

(c)

Compute the value of P (X < 8) as follows:

$P(X$

Thus, the value of P (X < 8) is 0.3707.

(d)

Compute the value of P (X < 20) as follows:

$P(X$

Thus, the value of P (X < 20) is 0.9525.

(e)

Compute the value of P (X > 16) as follows:

$P(X>16)=P(\frac{x-\mu}{\sigma}>\frac{16-10}{\sqrt{36}})\\=P(Z>1)\\=1-P(Z$

Thus, the value of P (X > 16) is 0.1587.

**Use a z-table for the probabilities.

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1 year ago

Dennis is numbering the pages of his book. Page 1 is right hand,page 2 a left hand page and faces 3,a right hand page. Which two

Aleks [24]

I drew it on paper and I got B as the pages that will face each other.

Answer: B. 22 and 23

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1 year ago

The blue M&M was introduced in 1995. Before then, the color mix in a bag of plain M&Ms was (30% Brown, 20% Yellow, 20% R

IRISSAK [1]

Answer:

The probability that the yellow M&M came from the 1994 bag is 0.07407 or 7.407%

Step-by-step explanation:

Given

Before 1995

(Br) Brown = 30%

(Y) Yellow = 20% =0.2

(R) Red = 20%

(G) Green =10% =0.1

(O) Orange = 10%

(T) Tan = 10%

After 1995

(Br) Brown = 13%

(Y) Yellow = 14% =0.14

(R) Red = 13%

(G) Green = 20% = 0.2

(O) Orange = 16%

(Bl) Blue = 24%

Since there are two bags, let A be the bag from 1994, and B be the bag from 1996

Then let AY imply we drew a yellow M&M from the 1994 bag

AG implies we drew a green M&M from the 1994 bag

BY implies imply we drew a yellow M&M from the 1996 bag

BG implies we drew a green M&M from the 1996 bag

P(AY) =0.2

P (BY) = 0.14

P(AG) =0.1

P(BG) =0.2

Since the draws from the 1994 and 1996 bag are independent,

therefore

$P(AY n BG) = 0.2 * 0.2 = 0.04 -------(1)\\P(AG n BY) =0.1 * 0.14 =0.014 --------(2)\\$

The draws can happen in either of the 2 ways in (1) and (2) above

therefore total probability E is given as

E = $P( AY n BG) u P(AG n BY)\\=0.04 + 0.014 =0.O54$

For the yellow one to be from 1994, it implies that the event to be chosen is

$P(AYnBG) = 0.2*0.2$

Since the total probability is given as E=0.054

then $P((AYnBG) /E) =\frac{0.04}{0.054} = 0.07407$

Concluding statement: This is the condition for the Yellow one to come from 1994 and green from 1996 provided that they obey the condition from E

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2 years ago

11/12−6/1q+6/5q−3/1, Combine like terms to create an equivalent expression.

Nikolay [14]

Answer: $-\frac{25}{12} -\frac{24q}{5}$

Step-by-step explanation:

$\frac{11}{12} -\frac{6}{1} q+\frac{6}{5}q-\frac{3}{1}$

$\frac{11}{12} -6q+ \frac{6q}{5} -\frac{3}{1}$

$\frac{11}{12}-6q \frac{6q}{5} -3$

$(\frac{11}{12} -3)+(-6q+\frac{6q}{5} )$

$-\frac{25}{12}- \frac{24q}{5}$

Simplify three times, collect like terms and simplify again, how to collect like terms see their differences and then put them into groups, for example I put all the numbers without variables on one side and all the numbers with variables on another.

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great rest of Black History Month! :-)

- Cutiepatutie ☺❀❤

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1 year ago

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