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
const2013 [10]
3 years ago
13

A set of data is described as:

Mathematics
2 answers:
Kisachek [45]3 years ago
8 0

Answer:

B

Step-by-step explanation:

saul85 [17]3 years ago
4 0

Answer:

b

Step-by-step explanation:

You might be interested in
Kenny wants to learn how to serve overhand in volleyball. When he practiced yesterday, he made 7 serves over the net for every 2
Yakvenalex [24]

Answer: 28

Step-by-step explanation:

Given

Kenny made 7 serves over the net for every 2 serves that did not go over the net i.e. success rate of Kenny is

\Rightarrow \text{success rate}=\dfrac{7}{2+7}\times 100=\dfrac{700}{9}\%

for 36 serves, applying the same success rate, it is

\Rightarrow \text{success rate}=\dfrac{700}{9}\times\dfrac{1}{100}\times 36=28

Serves that did not make over the net is 36-28=8

Thus, Kenny will make 28 serves that make over the net

8 0
3 years ago
What is -4R-2r+5 as a simpler exspression
enyata [817]
\sf-4R-2r+5 is already in simplest form.

But if you meant to say \sf-4r-2r+5, we would combine the first two terms.

Adding/subtracting like terms is the same as adding/subtracting whole numbers.

\sf-4-2=-6

Therefore:

\sf-4r-2r=-6r

Which gives us:

\boxed{\sf-6r+5}
8 0
3 years ago
Read 2 more answers
Please help . I’ll mark you as brainliest if correct !!!!
pav-90 [236]
31-21=15
Answer: 15m
Answer check: 36+15=51✅
3 0
3 years ago
Read 2 more answers
Micayla wants the print shop to reduce the size of her painting but keep the ratio of length to width so that it will fit her fr
marissa [1.9K]

Answer:

The width of the frame must be 12 inches (15in by 12in)

Step-by-step explanation:

Given

See attachment for painting and frame

Required

Determine the dimension of the frame

To do this, we make use of the following scale ratio.

Scale = Length : Width

For the painting,

Scale = 20: 16

For the frame,

Scale = 15: w

Since she wants to maintain the same ratio; both scale must be the same.

i.e.

15 : w = 20 : 16

Express as fraction

\frac{w}{15} = \frac{16}{20}

\frac{w}{15} = 0.8

Make w the subject

w = 15 * 0.8

w = 12

<em>The width of the frame must be 12 inches i.e. 15in by 12in frame</em>

8 0
3 years ago
The equation 2x + 4k − 9 = kx − k + 1 is really a family of equations, because for each value of k, we get a different equation
Olegator [25]
You haven't shared "the given value of x," or, if you have, you haven't drawn attention to it.

Just suppose we were to choose x = 4 as a possible solution and then try to find a value of the parameter k that would make x = 4 an actual solution.

2(4) + 4k - 9 = (4)(4) - (4) + 1

Then 8 + 4k - 9 = 16 - 4 + 1, or       4k - 1 = 13.  Then 4k = 14, and k = 14/4, or (after reduction)  k = 7/2   

If the parameter k equals 7/2, then x = 4 is a solution to the given equation.

To check this out further, start with the proposed solution x = 5 and find k.
3 0
3 years ago
Other questions:
  •   The mean is defined as the     
    8·2 answers
  • -
    10·1 answer
  • A bag of marbles contains 5 red, 3 blue, 2 green, and 2 yellow marbles. What is the probability that you choose a red marble and
    6·2 answers
  • A guy wire is attached to an upright pole 6 meters above the ground. If the wire is anchored to the ground 4 meters from the bas
    13·1 answer
  • Integral of sec (3x) tan (3x) dx
    7·1 answer
  • If f(a) = 3a? – a, then what is the value of f(-2) ?
    13·1 answer
  • Please help me with this one!!!
    10·2 answers
  • Byron is 2 years younger than Cindy. Eight
    8·2 answers
  • Im looking for salena
    12·2 answers
  • Find the domain and range of f(x). Please explain your answer.<br> f(x) = √(x^2+5x+6)
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!