Draw a circle at the coordinates listed

ABC and EDC are straight lines. EA is parallel to DB. EC = 8.1 cm. DC = 5.4 cm. DB = 2.6 cm. (a) Work out the length of AE. cm (

harkovskaia [24]

By applying the knowledge of similar triangles, the lengths of AE and AB are:

a. $\mathbf{AE = 3.9 $ cm}\\\\$

b. $\mathbf{AB = 2.05 $ cm} \\\\$

See the image in the attachment for the referred diagram.

The two triangles, triangle AEC and triangle BDC are similar triangles.
Therefore, the ratio of the corresponding sides of triangles AEC and BDC will be the same.

This implies that:

AC/BC = EC/DC = AE/DB

Given:

$EC = 8.1 $ cm\\\\DC = 5.4 $ cm\\\\DB = 2.6 cm\\\\AC = 6.15 $ cm$

a. Find the length of AE:

EC/DC = AE/DB

Plug in the values

$\frac{8.1}{5.4} = \frac{AE}{2.6}$

Cross multiply

$5.4 \times AE = 8.1 \times 2.6\\\\5.4 \times AE = 21.06$

Divide both sides by 5.4

$AE = \frac{21.06}{5.4} = 3.9 $ cm$

b. Find the length of AB:

$AB = AC - BC$

AC = 6.15 cm

To find BC, use AC/BC = EC/DC.

Plug in the values

$\frac{6.15}{BC} = \frac{8.1}{5.4}$

Cross multiply

$BC \times 8.1 = 6.15 \times 5.4\\\\BC = \frac{6.15 \times 5.4}{8.1} \\\\BC = 4.1$

Thus:

$AB = AC - BC$

Substitute

$AB = 6.15 - 4.1\\\\AB = 2.05 $ cm$

Therefore, by applying the knowledge of similar triangles, the lengths of AE and AB are:

a. $\mathbf{AE = 3.9 $ cm}\\\\$

b. $\mathbf{AB = 2.05 $ cm} \\\\$

Learn more here:

brainly.com/question/14327552

3 0

3 years ago

A horticulturist working for a large plant nursery is conducting experiments on the growth rate of a new shrub. Based on previou

valkas [14]

Answer:

$t=\frac{1.8-2}{\frac{0.5}{\sqrt{48}}}=-2.771$

The degrees of freedom are given by:

$df=n-1=48-1=47$

And the p value would be:

$p_v =P(t_{(47)}$

Since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case for the growth rate is less than 2cm per week

Step-by-step explanation:

Information given

$\bar X=1.8$ represent the sample mean for the growth

$s=0.5$ represent the sample standard deviation

$n=48$ sample size

$\mu_o =2$ represent the value that we want to compare

$\alpha=0.01$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is less than 2cm per week, the system of hypothesis are :

Null hypothesis: $\mu \geq 2$

Alternative hypothesis: $\mu < 2$

Since we don't know the population deviation the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{1.8-2}{\frac{0.5}{\sqrt{48}}}=-2.771$

The degrees of freedom are given by:

$df=n-1=48-1=47$

And the p value would be:

$p_v =P(t_{(47)}$

Since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case for the growth rate is less than 2cm per week

6 0

3 years ago

Mario sells insurance. His goal is to sell 3 types of insurance to each family on a block. He also wants to sell more than 27 po

qaws [65]

When we divide both sides of the inequality,

3x > 27 by 3, 3x(3) > 27(3)

we get x > 9, as the simplified form of the inequality.

Thus, giving us the answer of:

x > 9

8 0

3 years ago

A school administrator bought 6 desk sets. Each set consists of one table and one chair. He paid $48.50 for each chalr. If he pa

zlopas [31]

The answer is $62.00

Explanation: If you take $48.50 times 6 you get $291 now subtract $291 from $663 you will have an answer of $372 that is how much the total of tables came to now divide 372 by 6 for the number of tables and you will be left with 62

7 0

3 years ago

How do I solve this I don't know how to solve this i dont under stand this

k0ka [10]

Use Pythagoras theorem
${13}^{2} + {x}^{2} = {26}^{2} \\ { {x}^{2} = 507} \\ x = 22.52$

3 0

3 years ago