Hi you can help me plis

Find the value of sin 0 cos(90°- 0) + cos 0 sin(90°-0)

Step2247 [10]

Step-by-step explanation:

$sin \theta \: cos(90 \degree - \theta) + cos \theta \: sin(90 \degree - \theta) \\ = sin(\theta + 90 \degree - \theta) \\ = sin(90 \degree) \\ = 1$

5 0

4 years ago

NEED HELP WITH ALGEBRA 2 Solve each equation: 5^(3-2x)=5^(-x)

skad [1K]

Your answer would be x=3

8 0

3 years ago

Use the power-reducing formulas to rewrite the expression as an equivalent expression that does not contain powers of trigonome

Delicious77 [7]

Step-by-step explanation:

Sin^2 (x) * Cos^2 (x) = {[1 - cos (2x)]/2}*{[1 + cos (2x)]/2}

Sin^2 (x) * Cos^2 (x) =[ 1 - cos^2 (2x)]/4

Sin^2 (x) * Cos^2 (x) = (1/4) - (1/4) * cos^2 (2x)

Sin^2 (x) * Cos^2 (x) = (1/4) - (1/4) * {[1 + cos (2*2x)]/2}

Sin^2 (x) * Cos^2 (x) = (1/4) - (1/8) * [1 + cos (4x)]

Sin^2 (x) * Cos^2 (x) = (1/4) - (1/8) - (1/8)* [cos (4x)]

Sin^2 (x) * Cos^2 (x) = (1/8) - (1/8)* [cos (4x)]

6 0

3 years ago

From an airplane at an altitude of 1200 m, the angle of depression to a rock on the ground measures 28 degrees. Find the distanc

stellarik [79]

Answer:

300m

Step-by-step explanation:

3 0

3 years ago

Suppose that prices of a certain model of a new home are normally distributed with a mean of $150,000. Use the 68-95-99.7 rule t

levacccp [35]

Answer:

68% of buyers paid between $147,700 and $152,300.

Step-by-step explanation:

We are given that prices of a certain model of a new home are normally distributed with a mean of $150,000.

Use the 68-95-99.7 rule to find the percentage of buyers who paid between $147,700 and $152,300 if the standard deviation is $2300.

Let X = prices of a certain model of a new home

SO, X ~ Normal( $\mu=150,000 ,\sigma=2,300$ )

The z score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean price = $150,000

$\sigma$ = standard deviation = $2,300

Now, according to 68-95-99.7 rule;

Around 68% of the values in a normal distribution lies between $\mu-\sigma$ and $\mu-\sigma$ .

Around 95% of the values occur between $\mu-2\sigma$ and $\mu+2\sigma$ .

Around 99.7% of the values occur between $\mu-3\sigma$ and $\mu+3\sigma$ .

So, firstly we will find the z scores for both the values given;

Z = $\frac{X-\mu}{\sigma}$ = $\frac{147,700-150,000}{2,300}$ = -1

Z = $\frac{X-\mu}{\sigma}$ = $\frac{152,300-150,000}{2,300}$ = 1

This indicates that we are in the category of between $\mu-\sigma$ and $\mu-\sigma$ .

SO, this represents that percentage of buyers who paid between $147,700 and $152,300 is 68%.

7 0

4 years ago

Hi you can help me plis ​

Hi you can help me plis