Sinθ/cosθtanθ=1 how is this identity true?

Kevin is 4 times as old as Daniel and is also 6 years older than Daniel

MAXImum [283]

Answer:

Part a) Daniel's age is 2 years

Part b) Kevin's age is 8 years

Step-by-step explanation:

The question is

Part a) How old is Daniel?

Part b) How old is Kevin?

Let

x ----> Kevin's age

y ----> Daniel's age

we know that

$x=4y$ -----> equation A

$x=y+6$ ----> equation B

Equate equation A and equation B

$4y=y+6$

solve for y

$4y-y=6$

$3y=6$

$y=2$

therefore

Daniel's age is 2 years

Find the value of x

substitute the value of y in any of the two equations

$x=4(2)=8\ years$

$x=2+6=8\ years$

therefore

Kevin's age is 8 years

6 0

3 years ago

8 - 3x = 17 Solve the equation

777dan777 [17]

8-3x=17 substract 8
-8 -8
-3x=17-8
-3x=9 divide by -3
/-3 /-3
x=9/-3
x=-3 Solved for x

4 0

3 years ago

Can someone help me out? (Show work on how to solve)

belka [17]

Answer:

dghggjj

Step-by-step explanation:

s gm knfc JC. jz. jz. .xx nkcmxd I nt☺️️

8 0

3 years ago

Read 2 more answers

Prism A and Prism B have identical bases. The height of prism B is twice the height of prism A. Determine the relationship betwe

Artemon [7]

Area of a prism would 1/2xBxHxL

If prism B is twice twice the Height, then compared to A the area of B would be..

1/2xBx(2H)xL.. so prism B would have twice the volume of A.

Hope this helps

4 0

3 years ago

Read 2 more answers

The USDA conducted tests for salmonella in produce grown in California. In an independent sample of 252 cultures obtained from w

Marat540 [252]

Answer:

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There no sufficient evidence to show that the proportion of salmonella in the region’s water differs from the proportion of salmonella in the region’s wildlife

Step-by-step explanation:

From the question we are told that

The first sample size is $n_1 = 252$

The number that tested positive is $k_1 = 18$

The second sample size is $n_2 = 476$

The number that tested positive is $k_2 = 20$

The level of significance is $\alpha = 0.01$

Generally the first sample proportion is mathematically represented as

$\^ p _1 = \frac{k_1 }{ n_1 }$

=> $\^ p _1 = \frac{18 }{ 252 }$

=> $\^ p _1 = 0.071$

Generally the second sample proportion is mathematically represented as

$\^ p _2 = \frac{k_2 }{ n_2 }$

=> $\^ p _2 = \frac{20 }{ 476}$

=> $\^ p _2 = 0.042$

The null hypothesis is $H_o : p_1 - p_2 = 0$

The alternative hypothesis is $H_a : p_1 - p_2 \ne 0$

Generally the test statistics is mathematically represented

$z = \frac{ \^ p_1 - \^ p_2 - ( p_ 1 - p_2 )}{ \sqrt{\frac{\^ p_1 (1-\^ p_1)}{ n_1 } + \frac{\^ p_2 (1-\^ p_2)}{ n_2 } } }$

=> $z = \frac{ 0.071 - 0.042 - 0 }{ \sqrt{\frac{0.071 (1-0.071)}{ 252 } + \frac{0.042 (1-\^ 0.042)}{ 476 } } }$

=> $z = 1.56$

From the z table the area under the normal curve to the right corresponding to 1.56 is

$P(Z > 1.56 ) =0.05938$

Generally the p-value is mathematically represented as

$p-value = 2 * P(Z > 1.56 )$

=> $p-value = 2 * 0.05938$

=> $p-value = 0.1188$

From the value obtained we see that $p-value > \alpha$ hence

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There no sufficient evidence to show that the proportion of salmonella in the region’s water differs from the proportion of salmonella in the region’s wildlife

5 0

3 years ago

Sinθ/cosθtanθ=1how is this identity true?​

Sinθ/cosθtanθ=1how is this identity true?