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bogdanovich [222]
3 years ago
6

A 2-column table with 7 rows. Column 1 is labeled sample with entries 1, 2, 3, 4, 5, 6, 7. Column 2 is labeled Sample Mean with

entries 154, 121, 160, 135, 140, 134, 129. Looking at this table of sample means, which value is the best estimate of the mean of the population?
Mathematics
2 answers:
grandymaker [24]3 years ago
8 0

Answer:

(B)136

Step-by-step explanation:

Rom4ik [11]3 years ago
7 0

Answer:

(B)136

Step-by-step explanation:

Edg 2020 your welcome luv <3

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Pls help!!!!!!!!!!!!!!!!
Ludmilka [50]

Answer:

B

Step-by-step explanation:

they are similar and 7 / 5 = 21 / 15

5 0
3 years ago
Read 2 more answers
Is this set closed or not closed under the operation ? Negative integers under multiplication
Romashka [77]

The set is not closed

Example:

-4 * -3 = 12  this is not a negative integer

5 0
3 years ago
Four times the reciprocal of a number equal 2 times the reciprocal of 8
garik1379 [7]

Answer:

Step-by-step explanation:

Let the number be y & reciprocal simply means inverse of the number

4×1/y=2×1/8 solving for y

4/y=2/8; y = 16

5 0
3 years ago
Help! <br> Find the value of X and simplify completely.
Svetach [21]

Answer:

x = 8√5

Step-by-step explanation:

In the figure attached, there are three right angle triangles.

So, by Pythagoras theorem,

y²+ 4² = z²

y² + 16 = z²------(1)

Similarly x² = 16² + y²--------(2)

and (16 + 4)² = x² + z²

20² = x² + z²-------(3)

Now we put the value of z² from equation 1 to equation 3

20² = x² + y² + 16

x² + y² = 400 - 16

x² + y² = 384

y² = 384 - x²------(4)

Now we put the value of y² from equation 4 to equation 2

x² = 16² + 384 - x²

2x² = 256 + 384

2x² = 640

x² = 320

x = 8√5

Therefore x = 8√5 will be the answer.

8 0
3 years ago
Read 2 more answers
Prove that: (Sec A- cosec A)(1+ tan A+cot A) = tan A× sec A - cot A × cosec A
mash [69]

Answer:

Step-by-step explanation:

(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA

we take the LHS so here goes,

(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA\\(\frac{1}{cosA} -\frac{1}{sinA})(1+\frac{sinA}{cosA}+\frac{cosA}{sinA})\\\\(\frac{sinA-cosA}{sinAcosA})(\frac{sinAcosA+sin^2A+cos^2A}{sinAcosA})\\

since , sin^2A+cos^2A=1

the identity becomes,

(\frac{sinA-cosA}{sinAcosA})(\frac{1+sinAcosA}{sinAcosA})\\\\(\frac{sinA+sin^2AcosA-cosA-cos^2AsinA}{sin^2Acos^2A})\\\\

now, we know,

sin^2A=1-cos^2A and cos^2A=1-sin^2A

the identity becomes,

(\frac{sinA+(1-cos^2A)cosA-cosA-(1-sin^2A)sinA}{sin^2Acos^2A} )\\\\

(\frac{sinA+cosA-cos^3A-cosA-sinA+sin^3A}{sin^2Acos^2A})

sin A and cos A cancel out it becomes zero

\frac{sin^3A-cos^3A}{sin^2Acos^2A} \\\\

Splitting the denominator the identity becomes

\frac{sin^3A}{sin^2Acos^2A}-\frac{cos^3A}{sin^2Acos^2A}  \\\\\frac{sinA}{cos^2A} - \frac{cosA}{sin^2A} \\\\\frac{sinA}{cosA}(\frac{1}{cosA})-\frac{cosA}{sinA}(\frac{1}{sinA})\\\\

Hence,

tanAsecA-cotAcosecA

3 0
3 years ago
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