How high up does the ladder reach

Marta_Voda [28]

First, you would have to add d to both sides to get rid of it
x+d=ab+c/b
Then you would multiply b by both sides to get rid of the b in the denominator
x+d(b)=ab+c
After that, you would subtract c from both sides
x+d(b)-c=ab
Then you would divide both sides by a
x+d(b)-c=b
------------
a

6 0

3 years ago

A student records the number of hours that they have studied each of the last 23 days. They compute a sample mean of 2.3 hours a

natita [175]

Answer:

the standard deviation increases

Step-by-step explanation:

Let x₁ , x₂, . . . , x₂₃ be the actual data observed by the student

The sample means = x₁ + x₂ + . . . , x₂₃ / 23

$= \frac{x_1 +x_2 +...x_2_3}{23}$

= 2.3hr

⇒ $\sum xi =2.3 \times 23 = 52.9hrs$

let x₁ , x₂, . . . , x₂₃ arranged in ascending order

Then x₂₃ was 10 and has been changed to 14

i.e x₂₃ increase to 4

Sample mean $= \frac{x_1 +x_2 +...x_2_3}{23}$

$\frac{52.9hrs + 4}{23} \\\\= \frac{56.9}{23} \\\\= 2.47$

therefore, the new sample mean is 2.47

2) For the old data set

the median is $x_1_2(th)$ values

$[\frac{n +1}{2} ]^t^h value$

when we use the new data set only x₂₃ is changed to 14

i.e the rest all observation remain unchanged

Hence, sample median = $[{x_1_2]^t^h value$ remain unchange

sample median = 2.5hrs

The Standard deviation of old data set is calculated

$=\sqrt{\frac{1}{n-1} \sum (xi - \bar x_{old})^2 } \\\\=\sqrt{\frac{1}{22}\sum ( xi - 2.3)^2 }---(1)$

The new sample standard sample deviation is calculated as

$= \sqrt{\frac{1}{n-1} \sum (xi-2.47)^2} ---(2)$

Now, when we compare (1) and (2) the square distance between each observation xi and old mean is less than the squared distance between each observation xi and the new mean.

Since,

(xi - 2.3)² ∑ (xi - 2.47)²

Therefore , the standard deviation increases

6 0

3 years ago

If cade shoots 275 free throw baskets in 2 hours, how many can he shoot in 5 hours

Volgvan

Hey there! :D

Find the unit rate.

275/2= 137.5

Cade shoots 137.5 free throws per hour. Multiply that by 5 for 5 hours.

5*137.5= 687.5

He can shoot 687.5 free throws in one hour. You could just round that to 688.

I hope this helps!
~kaikers

6 0

3 years ago

Read 2 more answers

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

Work out the length of BC.

Akimi4 [234]

The length of BC is 53 units long

3 0

3 years ago