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3 years ago

A rectangle has dimensions of 12 cm long and 10 cm wide what is the area of the rectangle

1 answer:

3 0

The area and the width are known as 60 and 12 respectively. Rearranging that equation above, and plugging the values, one gets the value of the length of the rectangle as 5.

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Work out m and c for the line: y = x + 4

OLEGan [10]

Answer:

Step-by-step explanation:

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3 years ago

Which of the three runners have the fastest pace? Explain how you know.

GenaCL600 [577]

What is the image, i can’t see it?

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3 years ago

zubka84 [21]

Answer:

See Explanation Below

Step-by-step explanation:

Given

$(sin x - cos x)^2 = sec^2x - tan^2x - 2sinx.cos x.$

Required

Prove

To start with; we open the bracket on the left hand side

$(sin x - cos x)^2 = (sin x - cos x)(sin x - cos x)$

$(sin x - cos x)^2 = (sin x )(sin x - cos x)- (cos x)(sin x - cos x)$

$(sin x - cos x)^2 = sin^2 x -sinx cos x - sin xcos x + cos^2 x$

$(sin x - cos x)^2 = sin^2 x -2sinx cos x + cos^2 x$

Reorder

$(sin x - cos x)^2 = sin^2 x + cos^2 x - 2sinx cos x$

From trigonometry;

$sin^2x + cos^2x = 1$

So;

$(sin x - cos x)^2 = sin^2 x + cos^2 x - 2sinx cos x$

becomes

$(sin x - cos x)^2 = 1 - 2sinx cos x$

Also from trigonometry;

$sec^2x - tan^2x = 1$

So,

$(sin x - cos x)^2 = 1 - 2sinx cos x$

becomes

$(sin x - cos x)^2 = sec^2x - tan^2x - 2sinx cos x$

Proved

3 0

3 years ago

The total price for 7 movie tickets is 92.40 . If each ticket costs the same amount, how much does one movie ticket cost?

horsena [70]

Answer:

$13.20

Step-by-step explanation:

total cost of tickets = $92.40

total number of tickets = 7

cost of each ticket

= total cost of tickets ÷ total number of tickets

= 92.40 ÷ 7

= $13.20

hope this helps

8 0

3 years ago

Read 2 more answers

An airline finds that 5% of the persons making reservations on a certain flight will not show up for the flight. If the airline

vivado [14]

Answer:

The answer to the question is;

The probability that a seat will be available for every person holding a reservation and planning to fly is 0.63307.

Step-by-step explanation:

Let the sample size =n = 100

The success probability = 5 % = 0.05

Number of tickets sold = 105 tickets

In the case where there the airline has found that 5 % will not show up, then every passenger should have a seat, we have

A Binomial distribution is appropriate where there is a chance for a certain number of successful outcomes from a number of independent trails

However n·p and n·q must be ≥ 5 for there to be a normal approximation of a Binomial distribution thus

n·p = 105×0.05 = 5.25 ≥ 5

and n·q = n(1 - p) = 105 (1 - 0.05) = 99.75 ≥ 5

As the requirements are met, we can proceed with the approximation of the Binomial distribution by the normal distribution

$z = \frac{x-np}{\sqrt{np(1-p)} } = \frac{4.5 - 105*0.05}{\sqrt{105*0.05(1-0.05)} }$ = - 0.3358

We therefore have P(x ≥ 5) = P( x > 4.5) = P(z > -0.34) = 1 - P(z < -0.34) = 1 -0.36693 = 0.63307

Another way to solve the question is as follows

p = 0.95 q = 0.05

μ = np = 0.95*105 = 99.75, σ = $\sqrt{npq}$ = 2.233

P (x≤100) = P $(z$ = P(z<0.34) = 0.63307.

6 0

3 years ago