I'm so confused about conjecture. Not sure how to explain it to my daughter.

BARSIC [14]

A solution to a equation is when the variable is substituted fully. That makes the equation true.

3 0

3 years ago

A sport analyst wants to determine the mean salary of a Baseball player for 2015. He believes an estimate of this average salary

djyliett [7]

Answer:

The large sample size 'n' = 199.6569≅ 200

Step-by-step explanation:

Step(i):-

Given Standard deviation of salary for all baseball players for 2015 is about $5,478,384.55

Standard deviation of of salary for all baseball players for 2015

(S.D ) σ = $5,478,384.55

Given estimate of this average salary for all baseball players for 2015

= $497,000

Given Margin of error of error is = $497,000

Level of significance ∝ = 80%

The critical value Z₀.₂₀ = 1.282

Step(ii):-

Margin of error of error is determined by

$M.E = \frac{Z_{0.20} S.D}{\sqrt{n} }$

$497,000 = \frac{1.282 X 5,478,384.55}{\sqrt{n} }$

Cross multiplication , we get

$\sqrt{n} = \frac{1.282 X 5,478,384.55}{497,000 }$

On calculation , we get

√n = 14.13

Squaring on both sides, we get

n = 199.6569

Conclusion:-

The large sample size 'n' = 199.6569≅ 200

7 0

3 years ago

Solve 6 < x + 5 < 11

Vanyuwa [196]

Answer:

I think it should be 1 < x ≤ 6

Step-by-step explanation:

7 0

3 years ago

Pls helpppp ASAP I’m so confused

Sidana [21]

Answer:

2268 in3

Step-by-step explanation:

7 0

2 years ago

(cotx+cscx)/(sinx+tanx)

Butoxors [25]

Answer: $\bold{\dfrac{cot(x)}{sin(x)}}$

Step-by-step explanation:

Convert everything to "sin" and "cos" and then cancel out the common factors.

$\dfrac{cot(x)+csc(x)}{sin(x)+tan(x)}\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)}{1}+\dfrac{sin(x)}{cos(x)}\bigg)\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg[\dfrac{sin(x)}{1}\bigg(\dfrac{cos(x)}{cos(x)}\bigg)+\dfrac{sin(x)}{cos(x)}\bigg]\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)}{cos(x)}+\dfrac{sin(x)}{cos(x)}\bigg)$

$\text{Simplify:}\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)+sin(x)}{cos(x)}\bigg)\\\\\\\text{Multiply by the reciprocal (fraction rules)}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)cos(x)+sin(x)}\bigg)\\\\\\\text{Factor out the common term on the right side denominator}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)(cos(x)+1)}\bigg)$

$\text{Cross out the common factor of (cos(x) + 1) from the top and bottom}:\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)}\bigg)\\\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times cot(x)}\qquad \rightarrow \qquad \dfrac{cot(x)}{sin(x)}$

6 0

3 years ago