Multiply and simplify <br> tan θ(cos θ − csc θ)

Which object is 3 1/2 inches loong

Nastasia [14]

Answer:

ur mom

Step-by-step explanation:

expirence and funnes

8 0

2 years ago

5/3+3/8 fractions plz help me

sleet_krkn [62]

40/24+9/24= 40+9/24 = 49/24

8 0

3 years ago

Read 2 more answers

Assume that IQ scores are normally distributed with a mean of 100 and a standard deviation of 16. If 100 people are randomly sel

laiz [17]

Answer:

0.03 is the probability that for the sample mean IQ score is greater than 103.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 100

Standard Deviation, σ = 16

Sample size, n = 100

We are given that the distribution of IQ score is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$\dfrac{\sigma}{\sqrt{n}} = \dfrac{16}{\sqrt{100}} = 1.6$

P( mean IQ score is greater than 103)

P(x > 103)

$P( x > 103) = P( z > \displaystyle\frac{103 - 100}{1.6}) = P(z > 1.875)$

$= 1 - P(z \leq 1.875)$

Calculation the value from standard normal z table, we have,

$P(x > 103) = 1 - 0.970 =0.03= 3\%$

0.03 is the probability that for the sample mean IQ score is greater than 103.

7 0

3 years ago

Calculate the allele frequencies in this population of palm trees. A model consisting of a box containing fourteen, randomly arr

bogdanovich [222]

Answer:

The correct answer is

0.5 F, 0.5 f

Step-by-step explanation:

We note the following

Number of colored circles in the box = 14

Number of red circles in the box = 4

Number of purple circles in the box = 6

Number of blue circles in the box = 4

The allele frequency are as follows

Where the frequency is given as

Genotype Frequency Relative frequency

FF = Red 4 4/14 (0.29≈0.3)

Ff = Purple 6 6/14 (0.43≈0.4)

ff = Blue 4 4/14(0.29≈0.3)

Within this population,

We however have 4 FF = 8 F

6 Ff = 6 F + 6 f and

4 ff = 8 f

Total allele = 8+6+6+8 = 28

Relative frequency of F = (8+6)/28 = 14/28 = 0.5

relative frequency of f = (8+6)/28 = 0.5

Therefore the allele frequencies in the palm tree population is

0.5 F, 0.5 f

When in equilibrium we have

However the FF has the product of F×F which is = F² = 0.29 so the frequency of F = √(0.29) = 0.535≈ 0.5

The frequency of Ff is Ff or fF = 0.43 since there is equal number of each allele in Ff we have fF or Ff = Ff = 0.43

Which hives 0.43/2 = F =f ≈ 0.2

To

and ff = 0.29 so that f = 0.535 ≈ 0.5

Therefore f = F = 0.5 + 0.2 = 0.7

5 0

3 years ago

What is the equation of the line that passes through the points (−2, 1) and (1, 10)? 3x − y = −73x − y = −53x − y = 5x + 3y = −5

djyliett [7]

The two point formula :

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

The points are (-2, 1) and (1, 10)

Using the formula above :

$\begin{gathered} y-1=\frac{10-1}{1-(-2)}(x+2) \\ y-1=\frac{9}{3}(x+2) \\ y-1=3(x+2) \\ y-1=3x+6 \\ -3x+y=6+1 \\ -3x+y=7 \\ 3x-y=-7 \end{gathered}$

The answer is 3x - y = -7

7 0

1 year ago

Multiply and simplify tan θ(cos θ − csc θ)