Can u pls help me with this question asap

What is the answer to the equation 4x + - 9 = 74

zzz [600]

Answer:

x = 20.75

Step-by-step explanation:

<u>Step 1: Solve for x</u>

4x + (-9) = 74

4x - 9 + 9 = 74 + 9

4x / 4 = 83 / 4

x = 20.75

Answer: x = 20.75

3 0

3 years ago

Read 2 more answers

Stephen makes wooden shelves for his room he starts with a board that is 20 feet long.Stephen wants to make the shelves 3/4 foot

jek_recluse [69]

The number of shelves stephen can make is approximately 27

<h3><u>Solution:</u></h3>

Given that Stephen makes wooden shelves for his room

Stephen wants to make the shelves 3/4 foot long

He starts with a board that is 20 feet long

To find: number of shelves he can make

So the number of shelves stephen can make can make is found out by dividing 20 feet long board by 3/4 foot long

$\text {number of shelves }=\frac{\text { total length of board stephen has }}{\text { length of each board }}$

$\text {number of shelves }=\frac{20}{3 / 4}=\frac{20}{0.75}=26.6$

So the number of shelves stephen can make is approximately 27

3 0

3 years ago

For questions 13-15, Let Z1=2(cos(pi/5)+i Sin(pi/5)) And Z2=8(cos(7pi/6)+i Sin(7pi/6)). Calculate The Following Keeping Your Ans

weqwewe [10]

Answer:

Step-by-step explanation:

Given the following complex values Z₁=2(cos(π/5)+i Sin(πi/5)) And Z₂=8(cos(7π/6)+i Sin(7π/6)). We are to calculate the following complex numbers;

a) Z₁Z₂ = 2(cos(π/5)+i Sin(πi/5)) * 8(cos(7π/6)+i Sin(7π/6))

Z₁Z₂ = 18 {(cos(π/5)+i Sin(π/5))*(cos(7π/6)+i Sin(7π/6)) }

Z₁Z₂ = 18{cos(π/5)cos(7π/6) + icos(π/5)sin(7π/6)+i Sin(π/5)cos(7π/6)+i²Sin(π/5)Sin(7π/6)) }

since i² = -1

Z₁Z₂ = 18{cos(π/5)cos(7π/6) + icos(π/5)sin(7π/6)+i Sin(π/5)cos(7π/6)-Sin(π/5)Sin(7π/6)) }

Z₁Z₂ = 18{cos(π/5)cos(7π/6) -Sin(π/5)Sin(7π/6) + i(cos(π/5)sin(7π/6)+ Sin(π/5)cos(7π/6)) }

From trigonometry identity, cos(A+B) = cosAcosB - sinAsinB and sin(A+B) = sinAcosB + cosAsinB

The equation becomes

= 18{cos(π/5+7π/6) + isin(π/5+7π/6)) }

= 18{cos((6π+35π)/30) + isin(6π+35π)/30)) }

= 18{cos((41π)/30) + isin(41π)/30)) }

b) z2 value has already been given in polar form and it is equivalent to 8(cos(7pi/6)+i Sin(7pi/6))

c) for z1/z2 = 2(cos(pi/5)+i Sin(pi/5))/8(cos(7pi/6)+i Sin(7pi/6))

let A = pi/5 and B = 7pi/6

z1/z2 = 2(cos(A)+i Sin(A))/8(cos(B)+i Sin(B))

On rationalizing we will have;

= 2(cos(A)+i Sin(A))/8(cos(B)+i Sin(B)) * 8(cos(B)-i Sin(B))/8(cos(B)-i Sin(B))

= 16{cosAcosB-icosAsinB+isinAcosB-sinAsinB}/64{cos²B+sin²B}

= 16{cosAcosB-sinAsinB-i(cosAsinB-sinAcosB)}/64{cos²B+sin²B}

From trigonometry identity; cos²B+sin²B = 1

= 16{cos(A+ B)-i(sin(A+B)}/64

= 16{cos(pi/5+ 7pi/6)-i(sin(pi/5+7pi/6)}/64

= 16{ (cos 41π/30)-isin(41π/30)}/64

Z1/Z2 = (cos 41π/30)-isin(41π/30)/4

8 0

3 years ago

A teacher was interested in finding out whether a special study program would increase the scores of students on a national exam

bija089 [108]

Answer:

Check the explanation

Step-by-step explanation:

Going by the first attached image below we reject H_o against H_1 if obs. $T > t_{\alpha /2;n-1}$

here obs.T=1.879

$\therefore obs.T \ngtr 2.447=t_{0.025;6}$

we accept $H_o:\mu _{1}=\mu_{2}$ at 5% level of significance.

i.e there is no sufficient evidence to indicate that the special study program is more effective at 5% level of significance.

1.

this problem is simillar to the previous one except the alternative hypothesis.

Let X_i's denote the bonuses given by female managers and Y_i's denote the bonuses given by male managers.

we assume that $X_i \sim N(\mu _{1},\sigma _{1}^{2}) Y_i \sim N(\mu _{2},\sigma _{2}^{2})$ independently

We want to test $H_0:\mu_{1}=\mu_{2} vs H_1:\mu_{1}\neq \mu_{2}$

define $D_i=X_i-Y_i , i=1(1)8$

now $D_i\sim N(\mu _{1}-\mu _{2}=\mu _{D},\sigma _{1}^{2}+\sigma _{2}^{2}=\sigma _{D}^{2}) , i=1(1)8$

the hypothesis becomes

$H_0:\mu_{D}=0 vs H_1:\mu_{D}\neq 0$

in the third attached image, we use the same test statistic as before

i.e at 5% level of significance there is not enough evidence to indicate a difference in average bonuses .

3 0

3 years ago

Out of 100 what numbers are prime and composite

dalvyx [7]

Prime numbers 2 through 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97. Even numbers greater than two are composite.

5 0

3 years ago

Can u pls help me with this question asap ​

Can u pls help me with this question asap