What is 4x-5x= x- 20

Answers

Answer 1
Answer:

Answer: the answer is x=10

Answer 2
Answer:

Answer:   −2x + 20 = 0

Step-by-step explanation:


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Select the PROPER Scientific Notation for the following value. 659,000,000,000 Question 1 options: 659 × 109 65.9 × 1010 6.59 × 1011 0.659 × 1012

Answers

The proper notation may depend on where you are. In the US, we like there to be one non-zero digit to the left of the decimal point, hence the appropriate form is ...

... 6.59×10¹¹

_____

Elsewhere, the "proper" form may require the first non-zero digit be placed to the right of the decimal point, as in 0.659×10¹². This is also the form used by some computer languages.

In Engineering usage, quite often the desired form is one that has the exponent be a multiple of 3, so the appropriate forms would be 659×10⁹ or 0.659×10¹².

Sometimes when you're doing problems using numbers in scientific notation, it can be convenient to scale the operands so the result comes out with no need for rescaling. For example, if you were to divide 100 by this number, you might write ...

... 100/(65.9×10¹⁰) ≈ 1.517×10⁻¹⁰

Without this "pre-scaling", you would get

... (10²)/(6.59×10¹¹) ≈ 0.1517×10⁻⁹ = 1.517×10⁻¹⁰

which requires rescaling the answer.

Solve the matrix equation for a, b, c, and d. [1 2] [a b] [6 5][3 4] [c d]= [19 8]

Answers

Answer:

The answer is "\bold{\left[\begin{array}{cc}a&b\nc&d\end{array}\right] = \left[\begin{array}{cc}7&-2\n -(1)/(2)&(7)/(2)\end{array}\right]}".

Step-by-step explanation:

\bold{\left[\begin{array}{cc}1&2\n3&4\end{array}\right] \left[\begin{array}{cc}a&b\nc&d\end{array}\right] = \left[\begin{array}{cc}6&5\n 19&8\end{array}\right]}

Solve the L.H.S part:

\left[\begin{array}{cc}1&2\n3&4\end{array}\right] \left[\begin{array}{cc}a&b\nc&d\end{array}\right]\n\n\n\left[\begin{array}{cc}a+2c&b+2d\n3a+4c&3b+4d\end{array}\right]

After calculating the L.H.S part compare the value with R.H.S:

\left[\begin{array}{cc}a+2c&b+2d\n3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\n 19&8\end{array}\right]} \n\n

\to a+2c =6....(i)\n\n\to b+2d =5....(ii)\n\n\to 3a+4c =19....(iii)\n\n\to 3b+4d = 8 ....(iv)\n\n

In equation (i) multiply by 3 and subtract by equation (iii):

\to 3a+6c=18\n\to 3a+4c=19\n\n\text{subtract}... \n\n\to 2c = -1\n\n\to  c= - (1)/(2)

put the value of c in equation (i):

\to a+ 2 (- (1)/(2))=6\n\n\to a- 2 * (1)/(2)=6\n\n\to a- 1=6\n\n\to a =6 +1\n\n\to a = 7\n

In equation (ii) multiply by 3 then subtract by equation (iv):

\to 3b+6d=15\n\to 3b+4d=8\n\n\text{subtract...}\n\n\to 2d = 7\n\n\to d= (7)/(2)\n

put the value of d in equation (iv):

\to 3b+4 ((7)/(2))=8\n\n\to 3b+4 * (7)/(2)=8\n\n\to 3b+14=8\n\n\to 3b =8-14\n\n\to 3b = -6\n\n\to b= (-6)/(3)\n\n\to b= -2

The final answer is "\bold{\left[\begin{array}{cc}a&b\nc&d\end{array}\right] = \left[\begin{array}{cc}7&-2\n -(1)/(2)&(7)/(2)\end{array}\right]}".

One ticket is drawn at random from each of the two boxes.(A) 1 2 3 4 5

(B) 1 2 3 4 5 6

Find the chance that

(a) one of the number is 2 and the other is 5

(b) sum of the numbers is 7

(c) one number is bigger than twice the other

Answers

Answer:

a.1/15

b.1/6

c.1/3

Step-by-step explanation:

number of outcomes from box 1=5

number of outcomes from box 2=6

therefore total number of outcomes=6×5

=30

a. number of times receiving a ticket 2 and the other 5 =2

therefore probability= 2/30

=1/15

b.number of combinations for 7= 5 (1+6, 2+5,5+2,4+3,3+4.)

therefore probability= 5/30

=1/6

c.number of times one number was bigger than twice the other=10 (1+3,3+1,1+4,4+1,1+5,5+1,1+6,2+5,5+2,2+6)

therefore probability= 10/30

=1/3

The daily high temperature in Chicago for the month of August is approximately normal with mean 78 degrees F, and standard deviation 9 degrees F. a. What is the probability that a randomly selected day in August will have a high temperature greater than the mean daily high temperature of 78 degrees F?
b. What is the percentile for a day in August with a high temperature of 75 degrees F?
c. What is the 75th percentile for the daily high temperature for the month of August?
d. What is the interquartile range for the daily high temperature for the month of August?

Answers

Answer:

a) P(X>78) = P(Z> (78-78)/(9)) = P(Z>0)= 0.5

b) P(X<75)= P(Z< (75-78)/(9)) = P(Z<-0.333) = 0.370

So then 75 F correspond to approximately the 37 percentile

c) z=0.674<(a-78)/(9)

And if we solve for a we got

a=78 +0.674*9=84.07

So the value of height that separates the bottom 75% of data from the top 25% is 84.07 F.  

d) IQR = 84.07-71.93= 12.14

See explanation below.

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the daily high temperature in Chicago for the month of August of a population, and for this case we know the distribution for X is given by:

X \sim N(78,9)  

Where \mu=78 and \sigma=9

We are interested on this probability

P(X>78)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=(x-\mu)/(\sigma)

Using the z score we got:

P(X>78) = P(Z> (78-78)/(9)) = P(Z>0)= 0.5

Part b

For this case we can find the percentile with the following probability:

P(X<75)

If we use the z score formula we got:

P(X<75)= P(Z< (75-78)/(9)) = P(Z<-0.333) = 0.370

So then 75 F correspond to approximately the 37 percentile

Part c

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.25   (a)

P(X<a)=0.75   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.75 of the area on the left and 0.25 of the area on the right it's z=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=0.25

If we use condition (b) from previous we have this:

P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.75  

P(z<(a-\mu)/(\sigma))=0.75

But we know which value of z satisfy the previous equation so then we can do this:

z=0.674<(a-78)/(9)

And if we solve for a we got

a=78 +0.674*9=84.07

So the value of height that separates the bottom 75% of data from the top 25% is 84.07 F.  

Part d

For this case we know that IQR = Q_3 - Q_1 = P_(75)-P_(25)

So then we just need to find the percentile 25.

P(X>a)=0.25   (a)

P(X<a)=0.75   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

If we use condition (b) from previous we have this:

P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.25  

P(z<(a-\mu)/(\sigma))=0.25

But we know which value of z satisfy the previous equation so then we can do this:

z=-0.674<(a-78)/(9)

And if we solve for a we got

a=78 -0.674*9=71.93

So the value of height that separates the bottom 25% of data from the top 75% is 71.93 F.  

So then the interquartile range would be:

IQR = 84.07-71.93= 12.14

A person's heart rate is 65 beats per minute at rest. It increases by 12 beats per minute for each ofthe first 10 minutes of exercise. What is the person's heart rate after 6 minutes of exercise?

Answers

the heartbeat per minute is 137 i used calculator

Final answer:

The heart rate of the person after 6 minutes of exercise, assuming it increases by 12 beats per minute during this period, would be 137 beats per minute.

Explanation:

The question asks for the heart rate of a person after 6 minutes of exercise, given that it increases by 12 beats per minute during the first 10 minutes. To solve this, we start with the person's resting heart rate, which is 65 beats per minute. For each minute of exercise, we add 12 beats to the resting rate. So for 6 minute workout, the increase would be 6 times 12, which equals 72. Adding this to the resting heart rate, 65 + 72, the heart rate after 6 minutes of exercise would be 137 beats per minute.

Learn more about Heart Rate here:

brainly.com/question/14020895

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If your heart beats 65 times in one minute,How many times does it beat in one day?

Tips:
There are 60 minutes in one hour. There
Are 24 hours in a day

Answers

Answer:

93,600 times

Step-by-step explanation:

24x 60= 1445
in 1445 x 65 = 93,600
if your heart beats 65 times in 1445 minutes then it beats 93,600 in a day