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3 years ago

Please help problem 3 and 4show workfind the value of x, y and z

Mathematics

1 answer:

ratelena [41]3 years ago

3 0

Answer:

Step-by-step explanation:

The altitude to the hypotenuse of a right triangle create two smaller triangles, all of which are similar to the original. This means corresponding sides are proportional.

3. Using the above relationship, ...

short-side/hypotenuse = 8/y = y/(8+23)

y^2 = 8·31

y = 2√62

long-side/hypotenuse = z/(8+23) = 23/z

z^2 = 23·31

z = √713

short-side/long-side = 8/x = x/23

x^2 = 8·23

x = 2√46

_____

4. The picture is fuzzy, but we think the lengths are 25 and 5. If they're something else, use the appropriate numbers. Using the same relations we used for problem 3,

y = √(5·25) = 5√5 . . . . . . . = √(short segment × hypotenuse)

z = √(20·25) = 10√5 . . . . . = √(long segment × hypotenuse)

x = √(5·20) = 10 . . . . . . . . . = √(short segment × long segment)

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In right triangle DEF, it is known that Cos D = (12/13) and Cos F = (5/13). If FD = 39, then DE = ? Hint, draw the right triangl

AnnyKZ [126]

Answer:

The value of the line segment $ED$ is 36.

Step-by-step explanation:

The hypotenuse represents the longest side in the right triangle. In this case, FD represents the hypotenuse as it is a multiple of 13. Based on the trigonometric relations described in the statements, we get the following relationships by definition of cosines:

$\cos D = \frac{ED}{FD}$ (1)

$\cos F = \frac{EF}{FD}$ (2)

If we know that $\cos D = \frac{12}{13}$ and $FD = 39$ , then the length of the line segment $ED$ is:

$ED = FD\cdot \cos D$

$ED = 39\cdot \left(\frac{12}{13} \right)$

$ED = 36$

The value of the line segment $ED$ is 36.

7 0

2 years ago

Answer correctly / explain a little.

Dvinal [7]

Answer:

B) 8,4 pounds

Step-by-step explanation:

$y = 1,8[4] + 1,2; 8,4 = y \\ \: \: \: \: \: \: \: \: \: \: \: 7,2$

You substitute four weeks in for x, multiply that by 1,8 to get 7,2, and finally add 1,2 to it, to get the total weight loss of 8,4 pounds [3,81018 kilograms].

I am joyous to assist you anytime.

4 0

3 years ago

Read 2 more answers

Use the linear combination method to solve the system of equations.

laiz [17]

Answer:

Figures

figure 2.1

Figure 2.1

figure 2.10

Figure 2.10

figure 2.12

Figure 2.12

figure 2.13

Figure 2.13

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

The following table shows scores obtained in an examination by B.Ed JHS Specialism students. Use the information to answer the q

Makovka662 [10]

Answer:

(a) The cumulative frequency curve for the data is attached below.

(b) (i) The inter-quartile range is 10.08.

(b) (ii) The 70th percentile class scores is 0.

(b) (iii) the probability that a student scored at most 50 on the examination is 0.89.

Step-by-step explanation:

(a)

To make a cumulative frequency curve for the data first convert the class interval into continuous.

The cumulative frequencies are computed by summing the previous frequencies.

The cumulative frequency curve for the data is attached below.

(b)

(i)

The inter-quartile range is the difference between the third and the first quartile.

Compute the values of Q₁ and Q₃ as follows:

Q₁ is at the position:

$\frac{\sum f}{4}=\frac{100}{4}=25$

The class interval is: 34.5 - 39.5.

The formula of first quartile is:

$Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h$

Here,

l = lower limit of the class consisting value 25 = 34.5

(CF) $_{p}$ = cumulative frequency of the previous class = 24

f = frequency of the class interval = 20

h = width = 39.5 - 34.5 = 5

Then the value of first quartile is:

$Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h$

$=34.5+[\frac{25-24}{20}]\times5\\\\=34.5+0.25\\=34.75$

The value of first quartile is 34.75.

Q₃ is at the position:

$\frac{3\sum f}{4}=\frac{3\times100}{4}=75$

The class interval is: 44.5 - 49.5.

The formula of third quartile is:

$Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h$

Here,

l = lower limit of the class consisting value 75 = 44.5

(CF) $_{p}$ = cumulative frequency of the previous class = 74

f = frequency of the class interval = 15

h = width = 49.5 - 44.5 = 5

Then the value of third quartile is:

$Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h$

$=44.5+[\frac{75-74}{15}]\times5\\\\=44.5+0.33\\=44.83$

The value of third quartile is 44.83.

Then the inter-quartile range is:

$IQR = Q_{3}-Q_{1}$

$=44.83-34.75\\=10.08$

Thus, the inter-quartile range is 10.08.

(ii)

The maximum upper limit of the class intervals is 69.5.

That is the maximum percentile class score is 69.5th percentile.

So, the 70th percentile class scores is 0.

(iii)

Compute the probability that a student scored at most 50 on the examination as follows:

$P(\text{Score At most 50})=\frac{\text{Favorable number of cases}}{\text{Total number of cases}}$

$=\frac{10+4+10+20+30+15}{100}\\\\=\frac{89}{100}\\\\=0.89$

Thus, the probability that a student scored at most 50 on the examination is 0.89.

5 0

3 years ago

A random sample of 40 college students has a mean earnings of $3120 over the summer months. Assume the population standard devi

PSYCHO15rus [73]

Answer:

A normal distribution or z-test is used to construct a confidence interval.

Step-by-step explanation:

We are given the following in the question:

Sample mean, $\bar{x}$ = $3120

Sample size, n = 40

Population standard deviation, σ = $677

The distribution of earnings of college is a normal distribution.

Conditions:

Since we are given the population standard deviation and the the sample size is also greater than 30.

Conclusion:

Thus, we use a normal distribution or z-test to construct a confidence interval.

4 0

3 years ago

Please help problem 3 and 4show workfind the value of x, y and z​

Please help problem 3 and 4show workfind the value of x, y and z