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Phoenix [80]
3 years ago
15

Sandys friends ate 0, 2, 3, 4, 6, 6, and 7 pretzels. Sandy says the mean of the data is 4. Is Sandy correct?

Mathematics
2 answers:
diamong [38]3 years ago
7 0
If sandys friends ate 0, 2, 3, 4, 6, 6, and 7 pretzels, the mean of the data would be 4. So yes sandy is correct.
natka813 [3]3 years ago
4 0
Yes she is correct because if you add up all the numbers you get 28 then you divide that number by how many numbers there are so 7 and you get 4. 

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When solving proportions, we set the cross products equal and then we ________.
Olin [163]
Multiply each number across from another
6 0
3 years ago
What is the answer?
Amanda [17]
B seems right.
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4 0
3 years ago
Ming drew the model below to represent the equation 24+12=_x(8+4) What is the missing value in Ming's equation?
SashulF [63]

Given:

The equation is:

24+12=\_\_\times (8+4)

To find:

The missing value.

Solution:

We have,

24+12=\_\_\times (8+4)           ...(i)

Taking left hand side, we get

24+12=3\times 8+3\times 4

Taking 3 common, we get

24+12=3\times (8+4)             ...(ii)

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Missing value = 3

Therefore, the missing value in the given equation is 3.

4 0
2 years ago
I need help really bad please
Alina [70]

Answer:

Price of the hat = $14

Step-by-step explanation:

There are 20 bills, so $20.

He left $6. So,

Price of the hat = $20 - $6 = $14

6 0
2 years ago
Read 2 more answers
Bernoulli differential equation... y'+xy=xy^2
snow_lady [41]
y'+xy=xy^2\implies y^{-2}y'+xy^{-1}=x

Let z=y^{-1}, so that z'=-y^{-2}y'. Then the ODE becomes linear in z with

-z'+xz=x\implies z'-xz=-x

Find an integrating factor:

\mu(x)=\exp\left(\displaystyle\int-x\,\mathrm dx\right)=e^{-x^2/2}

Multiply both sides of the ODE by \mu:

e^{-x^2/2}z'-xe^{-x^2/2}z=-xe^{-x^2/2}

The left side can be consolidated as a derivative:

\left(e^{-x^2/2}z\right)'=-xe^{-x^2/2}

Integrate both sides with respect to x to get

e^{-x^2/2}z=e^{x^2/2}+C

where the right side can be computed with a simple substitution. Then

z=1+Ce^{x^2/2}

Back-substitute to solve for y.

y^{-1}=1+Ce^{x^2/2}\implies y=\dfrac1{1+Ce^{x^2/2}}
3 0
2 years ago
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