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2 years ago

What is the area of the kite?

Mathematics

1 answer:

enot [183]2 years ago

4 0

The area of the kite is 2.25 square feet

<h3>How to determine the area of the kite?</h3>

The area of the kite is calculated using:

Area = 0.5 * pq

Where p and q are the diagonals of the kite.

In this figure,

p = 3

q = 0.75 * 2 = 1.5

So, we have:

Area = 0.5 * 3 * 1.5

Evaluate

Area = 2.25

Hence, the area of the kite is 2.25

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Can someone solve these for me please

Grace [21]

Answer:

2. 16

3. 46

4. 59

Step-by-step explanation:

5.The measure of smaller angle is 87 degrees.

Step-by-step explanation:

Consider the provided information.

The provided angles are 3x degree and 2x + 25 degree.

Two angles are said to be Supplementary angles if their sum is 180 degree.

It is given that the provided angle are supplementary angles, thus

3x + 2x + 25 = 180

5x + 25 = 180

5x = 180 - 25

5x = 155

x = 31

Now, substitute the value of x in first angle i.e 3x.

3(31) = 93

Substitute the value of x in the second angle i.e 2x + 25

2(31) + 25 = 62 + 25 = 87

Hence, the measure of smaller angle is 87 degrees.

8 0

3 years ago

A new test to detect TB has been designed. It is estimated that 88% of people taking this test have the disease. The test detect

Elodia [21]

Answer:

Correct option: (a) 0.1452

Step-by-step explanation:

The new test designed for detecting TB is being analysed.

Denote the events as follows:

<em>D</em> = a person has the disease

<em>X</em> = the test is positive.

The information provided is:

$P(D)=0.88\\P(X|D)=0.97\\P(X^{c}|D^{c})=0.99$

Compute the probability that a person does not have the disease as follows:

$P(D^{c})=1-P(D)=1-0.88=0.12$

The probability of a person not having the disease is 0.12.

Compute the probability that a randomly selected person is tested negative but does have the disease as follows:

$P(X^{c}\cap D)=P(X^{c}|D)P(D)\\=[1-P(X|D)]\times P(D)\\=[1-0.97]\times 0.88\\=0.03\times 0.88\\=0.0264$

Compute the probability that a randomly selected person is tested negative but does not have the disease as follows:

$P(X^{c}\cap D^{c})=P(X^{c}|D^{c})P(D^{c})\\=[1-P(X|D)]\times{1- P(D)]\\=0.99\times 0.12\\=0.1188$

Compute the probability that a randomly selected person is tested negative as follows:

$P(X^{c})=P(X^{c}\cap D)+P(X^{c}\cap D^{c})$

$=0.0264+0.1188\\=0.1452$

Thus, the probability of the test indicating that the person does not have the disease is 0.1452.

4 0

3 years ago

How to simplify this expression??

sdas [7]

Answer :

Step-by-step-explanation :

${x}^{2} + 4x + 5 - {(x + 2)}^{2} \\ {x}^{2} + 4x + 5 - ( {x}^{2} + 4x + 4) \\$

${x}^{2} + 4x + 5 - {x}^{2} - 4x - 4 = {x}^{2} - {x}^{2} + 4x - 4x + 5 - 4 = 5 - 4 = 1$

8 0

3 years ago

Read 2 more answers

Mr. Jamison deposited $100 into a new savings account on January 1. On the first day of each month thereafter, he deposited thre

Lera25 [3.4K]

Answer:

$36 400

Step-by-step explanation:

Step 1

The first step is to figure out how much money is saved at the end of each month for the period from January 1 to June 15. The amount deposited at the end of each month is obtained by multiplying the amount from the previous month by 3.

The amount deposited in January is $\$100.$

The amount deposited in February is $1\$00\times 3= \$300.$

The amount deposited in March is $\$300\times 3= \$900.$

The amount deposited in April is $\$900\times 3= \$2\700.$

The amount deposited in May is $\$2\,700\times 3= \$8\,100.$

The amount deposited in June is $\$8\,100\times 3= \$24\,300.$

Step 2

The next step is to add up all the money that was deposited into the account. This calculation is shown below,

$\$100+\$300+900+\$\$2\,700+\$8\,100+\$24\,000=\$36\,400$

5 0

3 years ago

Read 2 more answers

Your deck is 3 2/3 feet off

Irina-Kira [14]

Your deck is bigger because it’s 3 and 2/3 and your neighbor is 3 and 8/9 your deck is bigger because 2/3 are bigger then 8/9

6 0

3 years ago

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