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3 years ago

Find the length of ad in the diagram shown

Mathematics

2 answers:

kvv77 [185]3 years ago

4 0

The answer to this is 7.5.

Ede4ka [16]3 years ago

4 0

Let as consider two triangles ABE and ACD

As we can see these two triangles ABE and ACD are similar triangles

Hence the ratio of sides must be equal

$\frac{AB}{AC} = \frac{AE}{AD}$

Now let AE = x

So plugging all the values we get

AB= 5 , AC= 11 , AE =x , AD = x+9

$\frac{5}{11} =\frac{x}{x+9}$

Cross multiplying we get

5(x+9) = 11(x)

Simplifying

5x +45 = 11x

Subtract 5x from both sides

45 = 6x

Divide both sides by 6

x= 45/6 = 7.5

Hence AE = 7.5

But we have to find the length of AD

AD = AE+ED = 7.5 + 9 = 16.5

AD= 16.5

Option D is the answer

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blondinia [14]

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3 years ago

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finlep [7]

Answer:

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Step-by-step explanation:

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3 0

3 years ago

In Everglades National Park in Florida, there are 200 species of birds that migrate. This accounts for 2/7 of all the species of

S_A_V [24]

Answer:

A) 2/7y=200

B) There are more species of plants as the number of species of birds that have been sighted in Everglades National Park is 280

Step-by-step explanation:

Total number of species of birds that migrate = 200

We are given that of all the species of birds are sighted in the park

a)Write an equation to find the number of species of birds that have been sighted in Everglades National Park.

Let y be the number of species of birds that have been sighted in Everglades National Park.

We are given that of all the species of birds are sighted in the park

So, 2/7y=200

y=200x7/4

y=280

B)There are 600 species of plants in Everglades National Park. Are there more species of birds or plants in the park?

So, There are more species of plants as the number of species of birds that have been sighted in Everglades National Park is 280

7 0

3 years ago

A kite is flying 11 ft off the ground. Its line is pulled taut and casts a 9 - ft shadow. Find the length of the line. If necess

jolli1 [7]

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8 0

3 years ago

Assume that women have heights that are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches.

Maurinko [17]

Answer:

Q3 = 65.7825.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 63.6, \sigma = 2.5$