1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
pshichka [43]
4 years ago
6

Solve the inequality, algebra nation part b

Mathematics
2 answers:
liraira [26]4 years ago
5 0
Solve the inequality, algebra nation part b

arsen [322]4 years ago
3 0
11 is answer i am nort sure
You might be interested in
If y varies inversely with x, find the constant of variation x= -40 and y= 4
Daniel [21]

Answer:

y=\frac{-160}{x}, \ \ \ \ k=-160

Step-by-step explanation:

let k be a constant written as:

y=\frac{k}{x}

It is given that y=4, x=-40

#We substitute x, y values in our function:

y=\frac{k}{x}\\\\4=\frac{k}{-40}\\\\k=4\times-40\\\\k=-160

We substitute the constant value in the equation:

y=\frac{k}{x}, k=-160\\\\y=\frac{-160}{x}

8 0
3 years ago
Joanne has a cylindrical, above ground pool. the depth (height) of the pool is 1/2 of its radius, and the volume is 1570 cubic f
Svetradugi [14.3K]

We know that Volume of Cylinder is given by : πr²h

Where : 'r' is the Radius of the Cylinder

             'h' is the Height or Depth of the Cylinder

Given : The Height of the Pool is Half of its Radius

⇒ Height of the Pool =  \frac{r}{2}

Given : The Volume of the Pool = 1570 feet³

⇒ πr²h = 1570

⇒ \pi (r^2)(\frac{r}{2}) = 1570

⇒ (\frac{22}{7})(\frac{r^3}{2}) = 1570

⇒ \frac{22r^3}{14} = 1570

⇒ r^3 = \frac{1570\times 14}{22} = 999

⇒ r = \sqrt[3]{999} = 10\;(approx)

As : Area of the Bottom of the Pool is Circular

We know that Area of Circle is given by : πr²

⇒ Area of the Bottom Floor = π × 10² = 314.15 feet²

5 0
3 years ago
How to factor a difference of squares over the integers.
nekit [7.7K]

Answer:

have a good day

Step-by-step explanation:

yeeeeeeeeet

4 0
3 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
(5x2 + 2x + 11) - (7 + 4x - 2x2)
storchak [24]
The answer is 7x^2 - 2x + 4, so answer D. 
6 0
3 years ago
Read 2 more answers
Other questions:
  • A boy ate 100 cookies in5 days. Each day he ate6 more than the day before. How many cookies did he eat the first day?
    11·1 answer
  • Suppose the primes p and q used in the rsa algorithm are consecutive primes. how would you factor n
    6·1 answer
  • Solve simultaneous equations using an appropriate method.
    14·2 answers
  • How do you check? <br> -5r+-3=51
    10·1 answer
  • Please helpp!!! Im putting 50
    5·2 answers
  • Each basket at the farm stand holds 3 pounds of potatoes or 6 pounds of tomatoes. If there are 21 pounds of potatoes and 6 pound
    15·2 answers
  • 3a²−4a+9=0 Use the discriminant to determine the number of real solutions to the quadratic equation.
    7·1 answer
  • 6,291 DIVIDED BY 3 = ? <br> its easy i know lets see if your smart.
    12·2 answers
  • A bag contains 9 discs of which 4 are red, 3 are blue and 2 are yellow. The discs are similar in shape and size. A disc is drawn
    8·1 answer
  • Graph the Quadratic equation y=x2+2x−3
    5·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!