PLEASE HELP

I need to know what the difference of this expression is. (4 - 5i) - (12 + 11i)

Juli2301 [7.4K]

Answer: -10-16i

Step-by-step explanation:

Given

(4-5i)-(12+11i)

Expand parenthesis

2-5i-12-11i

Put like terms together

(2-12)-(5i+11i)

Combine like terms

-10-16i

Hope this helps!! :)

Please let me know if you have any questions

8 0

3 years ago

A calzone is divided into 40 equal pieces.violet and ben ate 1/8 if the calzone on tuesday the next day ben ate 1/5 of the calzo

muminat

Answer:

28 pieces.

Step-by-step explanation:

1/8 of 40 is 5. Subtract that five (the five that they ate) from 40, leaving you with 35. It says the next day, Ben ate 1/5 of what was left. He ate 1/5 of 35 which is 7. He ate 7 pieces. Subtract this from 35 and you're left with 28 pieces of calzone.

5 0

4 years ago

(I have a deadline in 3 hours :c ) Ann's Coffee Shop makes a blend that is a mixture of two types of coffee. Type A coffee costs

max2010maxim [7]

A=3B
4.35A+5.40B=780.95
Substitute 3B into A for the second equation:
4.35(3B)+5.40B=780.90
13.05B+5.40B=780.90
18.45B=780.90
B=42.33
Plug into equation 1:
A=3B
A=3(42.33)
A=126.98

7 0

4 years ago

Read 2 more answers

Solve for x. Round to the nearest tenth of a degree, if necessary

NARA [144]

Answer:

29°

Step-by-step explanation:

cos x = 84/98

x = cos-1(84/98)

x = 28.9

5 0

3 years ago

You measure 50 textbooks' weights, and find they have a mean weight of 37 ounces. Assume the population standard deviation is 5.

Hoochie [10]

Answer:

90% confidence interval for the true population mean textbook weight is [35.79 ounces , 38.21 ounces].

Step-by-step explanation:

We are given that you measure 50 textbooks' weights, and find they have a mean weight of 37 ounces.

Assume the population standard deviation is 5.2 ounces.

Firstly, the Pivotal quantity for 90% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean weight = 37 ounces

$\sigma$ = population standard deviation = 5.2 ounces

n = sample of textbooks = 50

$\mu$ = true population mean textbook weight

Here for constructing 90% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5%

level of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X -1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X +1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X -1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X +1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $37-1.645 \times {\frac{5.2}{\sqrt{50} } }$ , $37+1.645 \times {\frac{5.2}{\sqrt{50} } }$ ]

= [35.79 , 38.21]

Therefore, 90% confidence interval for the true population mean textbook weight is [35.79 ounces , 38.21 ounces].

7 0

3 years ago