Convert 385 pounds into kg given the fact that 1 kg = 2.2 pounds.

What direction will this parabola open?

butalik [34]

Answer:

Step-by-step explanation:

4 0

2 years ago

A data set with a mean of 34 and a standard deviation of 2.5 is normally distributed

tresset_1 [31]

Answer:

a) $z= \frac{34-34}{2.5}= 0$

$z= \frac{39-34}{2.5}= 2$

And we want the probability from 0 to two deviations above the mean and we got 95/2 = 47.5 %

b) $P(X$

$z= \frac{31.5-34}{2.5}= -1$

So one deviation below the mean we have: (100-68)/2 = 16%

c) $z= \frac{29-34}{2.5}= -2$

$z= \frac{36.5-34}{2.5}= 1$

For this case below 2 deviation from the mean we have 2.5% and above 1 deviation from the mean we got 16% and then the percentage between -2 and 1 deviation above the mean we got: (100-16-2.5)% = 81.5%

Step-by-step explanation:

For this case we have a random variable with the following parameters:

$X \sim N(\mu = 34, \sigma=2.5)$

From the empirical rule we know that within one deviation from the mean we have 68% of the values, within two deviations we have 95% and within 3 deviations we have 99.7% of the data.

We want to find the following probability:

$P(34 < X$

We can find the number of deviation from the mean with the z score formula:

$z= \frac{X -\mu}{\sigma}$

And replacing we got

$z= \frac{34-34}{2.5}= 0$

$z= \frac{39-34}{2.5}= 2$

And we want the probability from 0 to two deviations above the mean and we got 95/2 = 47.5 %

For the second case:

$P(X$

$z= \frac{31.5-34}{2.5}= -1$

So one deviation below the mean we have: (100-68)/2 = 16%

For the third case:

$P(29 < X$

And replacing we got:

$z= \frac{29-34}{2.5}= -2$

$z= \frac{36.5-34}{2.5}= 1$

For this case below 2 deviation from the mean we have 2.5% and above 1 deviation from the mean we got 16% and then the percentage between -2 and 1 deviation above the mean we got: (100-16-2.5)% = 81.5%

7 0

3 years ago

For which system of equations is (5, 3) the solution? A. 3x – 2y = 9 3x + 2y = 14 B. x – y = –2 4x – 3y = 11 C. –2x – y = –13 x

Alla [95]

The correct answer is:

D) $\left \{ {{2x-y=7} \atop {2x+7y=31}} \right.$ .

Explanation:

We solve each system to find the correct answer.

For A:
$\left \{ {{3x-2y=9} \atop {3x+2y=14}} \right.$

Since we have the coefficients of both variables the same, we will use elimination to solve this.

Since the coefficients of y are -2 and 2, we can add the equations to solve, since -2+2=0 and cancels the y variable:
$\left \{ {{3x-2y=9} \atop {+(3x+2y=14)}} \right. 
\\
\\6x=23$

Next we divide both sides by 6:
6x/6 = 23/6
x = 23/6

This is not the x-coordinate of the answer we are looking for, so A is not correct.

For B:
$\left \{ {{x-y=-2} \atop {4x-3y=11}} \right.$

For this equation, it will be easier to isolate a variable and use substitution, since the coefficient of both x and y in the first equation is 1:
x-y=-2

Add y to both sides:
x-y+y=-2+y
x=-2+y

We now substitute this in place of x in the second equation:
4x-3y=11
4(-2+y)-3y=11

Using the distributive property, we have:
4(-2)+4(y)-3y=11
-8+4y-3y=11

Combining like terms, we have:
-8+y=11

Add 8 to each side:
-8+y+8=11+8
y=19

This is not the y-coordinate of the answer we're looking for, so B is not correct.

For C:
Since the coefficient of x in the second equation is 1, we will use substitution again.

x+2y=-11

To isolate x, subtract 2y from each side:
x+2y-2y=-11-2y
x=-11-2y

Now substitute this in place of x in the first equation:
-2x-y=-13
-2(-11-2y)-y=-13

Using the distributive property, we have:
-2(-11)-2(-2y)-y=-13
22+4y-y=-13

Combining like terms:
22+3y=-13

Subtract 22 from each side:
22+3y-22=-13-22
3y=-35

Divide both sides by 3:
3y/3 = -35/3
y = -35/3

This is not the y-coordinate of the answer we're looking for, so C is not correct.

For D:
Since the coefficients of x are the same in each equation, we will use elimination. We have 2x in each equation; to eliminate this, we will subtract, since 2x-2x=0:

$\left \{ {{2x-y=7} \atop {-(2x+7y=31)}} \right. 
\\
\\-8y=-24$

Divide both sides by -8:
-8y/-8 = -24/-8
y=3

The y-coordinate is correct; next we check the x-coordinate Substitute the value for y into the first equation:
2x-y=7
2x-3=7

Add 3 to each side:
2x-3+3=7+3
2x=10

Divide each side by 2:
2x/2=10/2
x=5

This gives us the x- and y-coordinate we need, so D is the correct answer.

7 0

3 years ago

Inductive Reasoning, help please :)

IgorLugansk [536]

Step-by-step explanation:

it may help you to understand

5 0

3 years ago

HELPP ME s + 3 Five6= 6 17

Oduvanchick [21]

Answer:The crystal structures of five 6-mercaptopurine derivatives, viz. 2-[(9-acetyl-9H-purin-6-yl)sulfanyl]-1-(3-methoxyphenyl)ethan-1-one (1), C16H14N4O3S, 2-[(9-acetyl-9H-purin-6-yl)sulfanyl]-1-(4-methoxyphenyl)ethan-1-one (2), C16H14N4O3S, 2-[(9-acetyl-9H-purin-6-yl)sulfanyl]-1-(4-chlorophenyl)ethan-1-one (3), C15H11ClN4O2S, 2-[(9-acetyl-9H-purin-6-yl)sulfanyl]-1-(4-bromophenyl)ethan-1-one (4), C15H11BrN4O2S, and 1-(3-methoxyphenyl)-2-[(9H-purin-6-yl)sulfanyl]ethan-1-one (5), C14H12N4O2S. Compounds (2), (3) and (4) are isomorphous and accordingly their molecular and supramolecular structures are similar. An analysis of the dihedral angles between the purine and exocyclic phenyl rings show that the molecules of (1) and (5) are essentially planar but that in the case of the three isomorphous compounds (2), (3) and (4), these rings are twisted by a dihedral angle of approximately 38°. With the exception of (1) all molecules are linked by weak C—H⋯O hydrogen bonds in their crystals. There is π–π stacking in all compounds. A Cambridge Structural Database search revealed the existence of 11 deposited compounds containing the 1-phenyl-2-sulfanylethanone scaffold; of these, only eight have a cyclic ring as substituent, the majority of these being heterocycles.

Keywords: crystal structure, mercaptopurines, supramolecular structure

Go to:

Chemical context

Purines, purine nucleosides and their analogs, are nitrogen-containing heterocycles ubiquitous in nature and present in biological systems like man, plants and marine organisms (Legraverend, 2008 ▸). These types of heterocycles take part of the core structure of guanine and adenine in nucleic acids (DNA and RNA) being involved in diverse in vivo catabolic and anabolic metabolic pathways.

6-Mercaptopurine is a water insoluble purine analogue, which attracted attention due to its antitumor and immunosuppressive properties. The drug is used, among others, in the treatment of rheumathologic disorders, cancer and prevent

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers