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marysya [2.9K]
3 years ago
7

I WILL GIVE BRAINLYEST IF IT"S RIGHT

Mathematics
2 answers:
Ierofanga [76]3 years ago
6 0

Answer:

B Sample space

Step-by-step explanation:

san4es73 [151]3 years ago
6 0
The correct answer to your question would be letter B!!!
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Use the metric ladder pictured above to complete the following conversions
V125BC [204]
50.0m = 5000cm
35.0m = 3500cm
8 0
3 years ago
7.) Solve the system of equations<br> below. <br> -2x + 3y = 9 2/3x + 2y = 0
iren [92.7K]

Answer:

So lets isolate y

we have

-2x+3y=9

2/3x +2y=0

Im gonna do the first equation

so -2x + 3y = 9

Add 2x on both sides

3y= 9+ 2x

Now divide by 3 to isolate y

y= 3+ 2/3x

Now we can subsitute y into any eqaution (I’m gonna do the second one) to solve.

2/3x+ 2(3+ 2/3x)= 0

Solve:

2/3x+ 6 + 4/3x = 0

6/3x +6= 0

2x+6=0

2x= -4

x= -2

Now we can subsitute this for any equation.

so I’m gonna do the first equation

-2(-2)+3y=9

4+3y=9

3y= 5

y= 5/3

y= 1 2/3

Final answer:

x=-2

y= 1 2/3

6 0
2 years ago
Use GeoGebra to explore the properties of rotations and complete each step belov
meriva

Answer:

This is the sample answer for edmentum.

5 0
3 years ago
dexamethasone 4 mg/1 ml is available. the MD orders 6000 mcg. how much will the nurse give? (Round to the tenth)​
aleksley [76]

Answer:

  1.5 mL

Step-by-step explanation:

The abbreviation mcg stands for "microgram," or 10^-6 grams. The abbreviation mg stands for "milligram," or 10^-3 grams. The relationship between these is ...

  1 mg = 1000 mcg

<h3>Dose</h3>

The order for 6000 mcg is an order for 6(1000 mcg) = 6(1 mg) = 6 mg. At 4 mg/mL, the quantity of dexamethasone required is ...

  (6 mg)/(4 mg/mL) = (6/4) mL = 1.5 mL

The nurse will give 1.5 milliliters of dexamethasone.

4 0
2 years ago
A sample of 18 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is n
ki77a [65]

Answer:

a)

i.  x=- - - - - -

ii.  o=- - - - - -

iii.  Sx=- - - - -

b) The random variable X signifies the weights of the bags of candies which are selected at random.  

c) The statistics variable X  is a measure on a sample that is used as an estimate of the population mean.X  is the mean weight of the 16 bags of candies that were selected.

d) The distribution needed for this problem is normal distribution with parameters  because the population standard deviation is known.

N(X, o/\sqrt{n})

So, the distribution is  

N(2, 0.12/\sqrt{16})

e)

i. The 90% confidence interval for the population mean weight of the candies is  1.9589 , 2.0411

iii. The error bound is 0.0411

f)

i. The 98% confidence interval for the population mean weight of the candies is  1.9418 , 2.0582

iii. The error bound is 0.0582

g) The difference in the confidence intervals in part (f) and part (e) is of the level of confidence. The change is, the change in the area being calculated for the normal distribution. Therefore, the larger confidence level results in larger area and larger interval.

Hence the interval in part  is larger than the one in part (e).

h) The interval in part (f) signifies that with 98% confidence that the population mean weight of the bag of candies lies between 1.942 ounce and 2.058 ounce.

Step-by-step explanation:

a)

i.  x=- - - - - -

ii.  o=- - - - - -

iii.  Sx=- - - - -

b) The random variable X signifies the weights of the bags of candies which are selected at random.  

c) The statistics variable X  is a measure on a sample that is used as an estimate of the population mean.X  is the mean weight of the 16 bags of candies that were selected.

d) The distribution needed for this problem is normal distribution with parameters  because the population standard deviation is known.

N(X, o/\sqrt{n})

So, the distribution is  

N(2, 0.12/\sqrt{16})

e)

i. The 90% confidence interval for the population mean weight of the candies is  1.9589 , 2.0411

iii. The error bound is 0.0411

f)

i. The 98% confidence interval for the population mean weight of the candies is  1.9418 , 2.0582

iii. The error bound is 0.0582

g) The difference in the confidence intervals in part (f) and part (e) is of the level of confidence. The change is, the change in the area being calculated for the normal distribution. Therefore, the larger confidence level results in larger area and larger interval.

Hence the interval in part  is larger than the one in part (e).

h) The interval in part (f) signifies that with 98% confidence that the population mean weight of the bag of candies lies between 1.942 ounce and 2.058 ounce.

5 0
3 years ago
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