Ask question

Ask question

All categories

3 years ago

7

David made 4 kites. Sthepen made three time as many kites as david.how many kites did the two boys make in all?

2 answers:

Zielflug [23.3K]3 years ago

6 0

Answer : 16 kites

David Made 4

Stephan Made : 3*4 = 12

Altogether the boys made : 16 kites

White raven [17]3 years ago

6 0

16 kites because 4 times 3 =12+4=16

You might be interested in

If kootie weighs 500 and gains 98 pounds in 2 days how much does he weigh

bezimeni [28]

598
explanation: 500+98= 598

5 0

2 years ago

Pemdaswhat is the PEMDAS of 40 - 32 / 8 + 5 x - 2

irina1246 [14]

Answer:

5x+34

Step-by-step explanation:

4 0

2 years ago

The mean radius of Main Street Park is 15 km and the mean radius of Uptown Park is 8 km. What is the approximate difference in t

Temka [501]

Step-by-step explanation:

circumference= 3.142*diameter

Diameter of main Street park is 30

Diameter of uptown park is 16

3.142*30=94.26

3.142*16=30.272

94.26-30.272=43.988

round off to tenth =44.0km

5 0

2 years ago

16) Graph g(x) = x + 3 + 2

krek1111 [17]

X=1.6 just did the math

3 0

1 year 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