All categories

3 years ago

Find the value of the variables in each figure. Explain your reasoning.

Mathematics

2 answers:

worty [1.4K]3 years ago

7 0

Answer: Number 7) x=14 and y=37

Number 8) x=28 and y=23

Step-by-step explanation: Number 7) Please see fig 1 in the attachment for naming of angle.

As per diagram line b||c and their transverse line a.

Using line and angle concepts.

{Corresponding angles of parallel line b||c}

{Supplementary angle of straight line}

Now combine the like term and solve for x and we get,

12x=180-12

x=14

For line a, 4y-10+3x=180 {Supplementary angle of straight line}

where, x=14

So, we get 4y-10+3(14)=180

4y=180-32

y=37

Number 8) Please see fig 2 in the attachment for naming of angles.

As per given diagram a||b||c and line d is transversal of all three line.

Using line and angle concepts.

{Corresponding angles of parallel line a||b}

{Supplementary angle of straight line}

3y+5y-4=180

8y=184

y=23

3y=2x+13 {Corresponding angles of parallel line b||c}

we calculated y=23 substitute into above equation and we get

3(23)=2x+13

2x=69-13

x=28

tiny-mole [99]3 years ago

5 0

Answer: ..................................................

You might be interested in

A manufacturer of chocolate candies uses machines to package candies as they move along a filling line. Although the packages ar

Elenna [48]

Answer:

We conclude that the mean amount packaged is equal to 8.17 ounces.

Step-by-step explanation:

We are given that in a particular sample of 50 packages, the mean amount dispensed is 8.171 ounces, with a sample standard deviation of 0.052 ounces.

Let $\mu$ = population mean amount packaged.

So, Null Hypothesis, $H_0$ : $\mu$ = 8.17 ounces {means that the mean amount packaged is equal to 8.17 ounces}

Alternate Hypothesis, $H_A$ : $\mu\neq$ 8.17 ounces {means that the mean amount packaged is different from 8.17 ounces}

The test statistics that will be used here is One-sample t-test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean amount dispensed = 8.171 ounces

s = sample standard deviation = 0.052 ounces

n = sample of packages = 50

So, the test statistics = $\frac{8.171-8.17}{\frac{0.052}{\sqrt{50} } }$ ~ $t_4_9$

= 0.1359

The value of t-test statistics is 0.1359.

Also, the P-value of test-statistics is given by;

the meaning of the p-value is that the p-value is the probability of obtaining a sample mean that is equal to or more extreme than 0.001 ounces away from8.17 if the null hypothesis is true.

P-value = P( $t_4_9$ > 0.136) = More than 40% {from the t-table}

Since the P-value of our test statistics is more than the level of significance of 0.01, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the mean amount packaged is equal to 8.17 ounces.

6 0

3 years ago

On Saturday, you spend $33, give $15 to a friend, and receive $20 for mowing your neighbor’s lawn. You have $21 left. Use two me

tia_tia [17]

21+20-15-33=x

x=-7

Hope this helped☺☺

8 0

3 years ago

Plz answer as fast possible

My name is Ann [436]

9. 500 5
6000 60
30000 300
10. 48 0.48
9.7 0.097
510.3 5.103

6 0