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Gnesinka [82]
3 years ago
15

Find the volume of the prism.

Mathematics
1 answer:
zubka84 [21]3 years ago
4 0

Answer:

b I do believe good luck with the answer

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A standard American Eskimo dog has a mean weight of 30 pounds with a standard deviation of 2 pounds. Assuming the weights of sta
lions [1.4K]

Answer:

Approximately 24–36 pounds

Step-by-step explanation:

Given:

Weight of standard American Eskimo dog has mean= 30 pounds

A standard deviation= 2 pounds

For 99.7% of the dogs according to normal distribution would lie between 3 standard deviations on either side of the mean.

i.e. lower bound= Mean- 3 (standard deviation)

= 30-3(2)

=30-6

=24

Upper bound = Mean +3 (standard deviation)

= 30+3(2)

=30+6

=36

Hence the range of weights : 24-36 pounds

correct answer Approximately 24–36 pounds!

4 0
3 years ago
A manager at a local company asked his employees how many times they had given blood in the last year. The results of the survey
Lubov Fominskaja [6]

Answer:

Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779

And the deviation would be:

Sd(X) =\sqrt{2.3779}= 1.542 \approx 1.54

Step-by-step explanation:

For this case we have the following distribution given:

X        0         1       2       3       4         5        6

P(X)  0.3   0.25   0.2   0.12   0.07   0.04   0.02

For this case we need to find first the expected value given by:

E(X) = \sum_{i=1}^n X_i P(X_I)

And replacing we got:

E(X)= 0*0.3 +1*0.25 +2*0.2 +3*0.12 +4*0.07+ 5*0.04 +6*0.02=1.61

Now we can find the second moment given by:

E(X^2) =\sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2)= 0^2*0.3 +1^2*0.25 +2^2*0.2 +3^2*0.12 +4^2*0.07+ 5^2*0.04 +6^2*0.02=4.97

And the variance would be given by:

Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779

And the deviation would be:

Sd(X) =\sqrt{2.3779}= 1.542 \approx 1.54

 

8 0
3 years ago
16. Solve for C: 6c - 1 - 4c = -49
lesantik [10]

Answer:

c = -24

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtract Property of Equality

<u>Algebra I</u>

<u />

Step-by-step explanation:

<u>Step 1: Define</u>

6c - 1 - 4c = -49

<u>Step 2: Solve for </u><em><u>c</u></em>

  1. Combine like terms:                     2c - 1 = -49
  2. Isolate <em>c</em> term:                               2c = -48
  3. Isolate <em>c</em>:                                        c = -24
7 0
2 years ago
For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4
Elina [12.6K]

Answer:

<em>C.</em> (5-\frac{1}{2})^6

Step-by-step explanation:

Given

15(5)^2(-\frac{1}{2})^4

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

Sum = 2 + 4

Sum = 6

Each term of a binomial expansion are always of the form:

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

Where n = the sum above

n = 6

Compare 15(5)^2(-\frac{1}{2})^4 to the above general form of binomial expansion

(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......

Substitute 6 for n

(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......

[Next is to solve for a and b]

<em>From the above expression, the power of (5) is 2</em>

<em>Express 2 as 6 - 4</em>

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

By direct comparison of

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

and

(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......

We have;

^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4

Further comparison gives

^nC_r = 15

a^{n-r} =(5)^{6-4}

b^r= (-\frac{1}{2})^4

[Solving for a]

By direct comparison of a^{n-r} =(5)^{6-4}

a = 5

n = 6

r = 4

[Solving for b]

By direct comparison of b^r= (-\frac{1}{2})^4

r = 4

b = \frac{-1}{2}

Substitute values for a, b, n and r in

(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......

(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......

Solve for ^6C_4

(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......

(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......

<em>Check the list of options for the expression on the left hand side</em>

<em>The correct answer is </em>(5-\frac{1}{2})^6<em />

3 0
3 years ago
What is the midpoint of a segment with These endpoints (-12,-3) and (3,-8)
Savatey [412]

Answer:

(\frac{-9}{2},\frac{11}{2})

Step-by-step explanation:

you have to use the midpoint formula which is

(\frac{x1+x2}{2},\frac{y1+y2}{2})

so you plug your numbers in

(\frac{-12+3}{2},\frac{-3+-8}{2})

Add the tops

(\frac{-9}{2} ,\frac{-11}{2} )

if possible simplify

and that is your answer

8 0
2 years ago
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