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3 years ago

!!!!How many solutions does the equation have?

Mathematics

2 answers:

Margaret [11]3 years ago

3 0

No solution bc the equations aren't equal to each other

aalyn [17]3 years ago

3 0

Answer:

one solution since it's a lender equation

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Identify the following number as rational or irrational 3.2573820193748...

Bingel [31]

Answer: Irrational

Step-by-step explanation:

4 0

3 years ago

(1+2i) (2+5i)

Lostsunrise [7]

Answer:

Step-by-step explanation:

first multiply 1 with 2+5i then multiply 2i with 2+5i and you will get

2+5i+4i+10i^2

then in the next step add 4i and 5i you will get 9i

in the next step put i^2=-1 and you will get -10

in the last step just substract 2-10 you will get -8 and 9i

and your answer will be -8+9i

5 0

4 years ago

A magazine conducts an annual survey in which readers rate their favorite cruise ship. All ships are rated on a 100-point scale,

Lyrx [107]

Answer:

a) The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$

b) $SE=\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=0.975$

And the Margin of error is given by:

$Me= z_{\alpha/2} * SE=1.96*0.975=1.910$

c) The 95% confidence interval would be given by $2.009 \leq \mu_1 -\mu_2 \leq 5.83$ .

Step-by-step explanation:

Notation and previous concepts

$n_1 =35$ represent the sample of ships that carry fewer than 500 passengers

$n_2 =44$ represent the sample of ships that carry 500 or more passengers

$\bar x_1 =85.82$ represent the mean sample of of ships that carry fewer than 500 passengers

$\bar x_2 =81.90$ represent the mean sample of of ships that carry 500 or more passengers

$\sigma_1 =4.55$ represent the population deviation of ships that carry fewer than 500 passengers

$\sigma_2 =3.97$ represent the sample deviation of ships that carry 500 or more passengers

$\alpha=0.05$ represent the significance level

Confidence =95% or 0.95

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{(\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2})}$ (1)

Part a

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$

Part b: At 95% confidence, what is the margin of error?

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

The standard error is given by the following formula:

$SE=\sqrt{(\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2})}$

And replacing we have:

$SE=\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=0.975$

And the Margin of error is given by:

$Me= z_{\alpha/2} * SE=1.96*0.975=1.910$

Part c: What is a 95% confidence interval estimate of the difference between the population mean ratings for the two sizes of ships?

Confidence interval

Now we have everything in order to replace into formula (1):

$3.92-1.96\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=2.009$

$3.92+1.96\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=5.830$

So on this case the 95% confidence interval would be given by $2.009 \leq \mu_1 -\mu_2 \leq 5.83$ .

7 0

3 years ago

An investment firm invested in two companies last year. They invested $18,000 in Company A and made a profit of 12%. They invest

VMariaS [17]

Answer: can somebody help me w my recent marking brainlist

Step-by-step explanation:

Sorry ik this isn’t the answer :(

6 0

3 years ago

Read 2 more answers

can someone please help me with this I'm failing this class and I need help please I'll mark you brainliest.

Murrr4er [49]

Answer:

1. 36 $m^{2}$

2. 333.125 $ft^{3}$

3. 210 $ft^{3}$

4. Yes, Martina's estimate is reasonable.

Step-by-step explanation:

Numbering the rectangular surfaces, we have:

Area of a rectangle = length x width

For surface 1: Area = 2 x 2.4

= 4.8 $m^{2}$

For surface 2: Area = 3 x 2.4

= 7.2 $m^{2}$

For surface 3: Area = 3 x 2

= 6.0 $m^{2}$

For surface 4: Area = 3 x 2.4

= 7.2 $m^{2}$

For surface 5: Area = 2 x 2.4

= 4.8 $m^{2}$

For surface 6: Area = 3 x 2

= 6.0 $m^{2}$

Total surface area of the rectangular prism = (2 x 4.8) + (2 x 7.2) + (2 x 6.0)

= 36 $m^{2}$

2. length = 10.25 ft

width = 5 ft

height = 6.5 ft

Thus,

volume = l x w x h

= 10.25 x 5 x 6.5

= 333.125 $ft^{3}$

3. length = 15 ft

width = 7 ft

height = 2 ft

So that,

volume = l x w x h

= 15 x 7 x 2

= 210 $ft^{3}$

4. For a rectangular prism, area of the opposite surfaces are equal. So that;

Area of rectangle = length x width

For surface 1: Area = 13.0 x 6

= 78.0 $ft^{2}$

For surface 2: Area = 13.0 x 8

= 104.0 $ft^{2}$

For surface 3: Area = 6 x 8

= 48 $ft^{2}$

Surface area of the prism = (2 x 78.0) + (2 x 104.0) + (2 x 48)

= 460.0 $ft^{2}$

Therefore, Martina's estimate is reasonable.

5 0

3 years ago