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2 years ago

X-3/18=12/9 what is the answer? It is proportions.

Mathematics

2 answers:

olga nikolaevna [1]2 years ago

8 0

Answer:

x=3/2

Step-by-step explanation:

qaws [65]2 years ago

8 0

Answer:

x= $\frac{3}{2}$

Step-by-step explanation:

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Answer 29 to 33......

Sindrei [870]

2y=x
30:174.24
31:25
32:2
33:16
hope this helps!!!!!

6 0

2 years ago

Use the circle graph to determine how many hours per day Becky spends on each activity.

Oduvanchick [21]

404 error: graph not found

anyway, the graph was not included. since the sleep time was included, I will assume that the circle graph is worth 24 hours

all we need to do is to convert the percentages to fractions and multiply that by 24 to find out how many hours per activity

percent means parts out of 100 so x%=x/100

so we have
School
Eating
Sleep
Homework
Free Time

School=25%
25%=25/100=1/4
1/4 times 24=6
School: 6 hours

Eating=10%
10%=10/100=1/10
1/10 times 24=2.4
Eating: 2.4 hours

Sleep=40%
40%=40/100=4/10=2/5
2/5 times 24=48/5=9.6
Sleep: 9.6 hours

Homework=10%
10%=10/100=1/10
1/10 times 24=2.4
Homework: 2.4 hours

Free Time=15%
15%=15/100=3/30
3/20 times 24=72/20=36/10=3.6
Free Time: 3.6 Hours

Answers:
School: 6 hours
Eating: 2.4 hours
Sleep: 9.6 hours
Homework: 2.4 hours
Free Time: 3.6 Hours

3 0

3 years ago

On monday 254 students went on a trip to the zoo all 5 buses were filled 9 students had to travel in cars how many students were

Nady [450]

254-9=245 then you do 245/5 which is equal to 49.

4 0

3 years ago

Read 2 more answers

An animal shelter spends $3.00 per day to care for each cat and $6.50 per day to care for each dog. Amelia noticed that the shel

Ede4ka [16]

Answer:

There were 15 cats at the shelter on Friday.

Step-by-step explanation:

Cost of caring each cat = $3.00

Cost for caring each dog = $6.50

Total money spent by shelter = $84.00

Total Cats and Dogs in the shelter = 21

Let the number of cats in the shelter = k

So, the number of dogs in shelter = 21 -k

Also according to the question:

Total expenditure on cats + total expenditure on dogs = $84.00

or, k ($3.00) + (21-k)($6.50) = $84.00

or, 3k + 136.5 - 6.5 k = 84

or, -3.5 k = 84 - 136.5 = -52.5

⇒3.5 k = 52.5 or k = 52.5/3.5 = 15

⇒ k = 15

Hence, the number of cats in the shelter was 15 on Friday.

7 0

3 years ago

When the author visited Dublin, Ireland (home of Guinness Brewery employee William Gosset, who first developed the t distributio

disa [49]

Answer:

The Decision Rule

Fail to reject the null hypothesis

The conclusion

There is no sufficient evidence to support the claim that the mean age of the cars is greater than that of taxi

Step-by-step explanation:

From the question we are told that

The data is

Car Ages 4 0 8 11 14 3 4 4 3 5 8 3 3 7 4 6 6 1 8 2 15 11 4 1 6 1 8

Taxi Ages 8 8 0 3 8 4 3 3 6 11 7 7 6 9 5 10 8 4 3 4

The level of significance $\alpha = 0.05$

Generally the null hypothesis is $H_o : \mu_1 - \mu_2 = 0$

the alternative hypothesis is $H_a : \mu_1 - \mu_2 > 0$

Generally the sample mean for the age of cars is mathematically represented as

$\= x_1 = \frac{\sum x_i }{n}$

=> $\= x_1 = \frac{4+ 0+ 8 +11 + \cdots + 8
}{27}$

=> $\= x_1 = 5.56$

Generally the standard deviation of age of cars

$\sigma _1 = \sqrt{\frac{\sum (x_i - \= x)^2}{n_1} }$

=> $\sigma _1 = \sqrt{\frac{(4 - 5.56)^2 + (0 - 5.56)^2+ (8 - 5.56)^2 + \cdots + 8}{ 27} }$

=> $\sigma _1 = 3.88$

Generally the sample mean for the age of taxi is mathematically represented as

$\= x_2 = \frac{\sum x_i }{n}$

=> $\= x_2 = \frac{8 +8 +0 + \cdots + 4
}{20}$

=> $\= x_2 = 5.85$

Generally the standard deviation of age of taxi

$\sigma _2 = \sqrt{\frac{\sum (x_i - \= x)^2}{n_1} }$

=> $\sigma _2 = \sqrt{\frac{(8 - 5.85)^2 + (8 - 5.85)^2+ (0 - 5.85)^2 + \cdots + 8}{ 20} }$

=> $\sigma _2 = 2.83$

Generally the test statistics is mathematically represented as

$t = \frac{(\= x_ 1 - \= x_2 ) - 0}{\sqrt{\frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2} } }$

=> $t = \frac{(5.56 - 5.85 ) - 0}{\sqrt{\frac{(3.88)^2}{27} + \frac{(2.83)}{20} }}$

=> $t = -0.30$

Generally the degree of freedom is mathematically represented as

$df = n_1 + n_2 -2$

$df = 27 + 20 -2$

$df = 45$

From the t distribution table the $P(t > t )$ at the obtained degree of freedom = 45 is

$P(t > -0.30 ) = 0.61722067$

So the p-value is

$p-value = P(t > T) = 0.61722067$

From the obtained values we see that the p-value > $\alpha$ hence we fail to reject the null hypothesis

Hence the there is no sufficient evidence to support the claim that the mean age of the cars is greater than that of taxi

5 0

3 years ago