Find 75 from 7 6 5 3

Help please please 100pts no trolls please

ycow [4]

Answer:

21.8 m/s

Step-by-step explanation:

Given that Tom drops a object from a height of h meters and it hits the ground with a veloc8of v m/s and it is given by function ;

$\longrightarrow v =\sqrt{19.6h}$

We need to find out the velocity upon reaching the ground when it is dropped from a height of 24.3 m . On plugging in 24.3 in place of h we have ,

$\longrightarrow v =\sqrt{19.6\times 24.3 }\ m/s\\\\\longrightarrow v =\sqrt{ 476.28} m/s \\\\\longrightarrow v = 21.823 \ m/s \\\\\longrightarrow \underline{\underline{ v = 21.8 \ m/s }}$

And we are done !

8 0

3 years ago

Read 2 more answers

Angie takes a random sample of 100 students in her school and finds that 58% of the sample prefers art over music. There are 120

REY [17]

The answer is
$a = 696$
because

$1200 \div 100 \times 58 = 696$

8 0

3 years ago

Read 2 more answers

A person drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed f

Olin [163]

Answer:

$\chi^2 = \frac{(26-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33}+\frac{(44-33.33)^2}{33.33}+\frac{(37-33.33)^2}{33.33}+\frac{(27-33.33)^2}{33.33}+\frac{(34-33.33)^2}{33.33}=6.7$

Now we can calculate the degrees of freedom for the statistic given by:

$df=6-1=5$

And we can calculate the p value given by:

$p_v = P(\chi^2_{5} >6.7)=0.244$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we can conclude that the outcomes are equally likely

Step-by-step explanation:

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference in the frequencies

H1: There is a difference in the frequencies

The level of significance assumed for this case is $\alpha=0.01$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The observed values are:

Value 1 2 3 4 5 6

Frequency 26 32 44 37 27 34

And the expected values are for this case the same $E_i = \frac{200}{6}= 33.33$

And now we can calculate the statistic:

$\chi^2 = \frac{(26-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33}+\frac{(44-33.33)^2}{33.33}+\frac{(37-33.33)^2}{33.33}+\frac{(27-33.33)^2}{33.33}+\frac{(34-33.33)^2}{33.33}=6.7$

Now we can calculate the degrees of freedom for the statistic given by:

$df=6-1=5$

And we can calculate the p value given by:

$p_v = P(\chi^2_{5} >6.7)=0.244$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we can conclude that the outcomes are equally likely

7 0

4 years ago

Help with this ( I have trouble with this) make sure to round it to the nearest hundredth as needed

8090 [49]

9^3= 729, 729 x 4=2916, 2916/3=972, 972 x 3.14 =3052.08

3 0

3 years ago

Read 2 more answers

Solve for x can someone please answer help

lakkis [162]

Answer:

Step-by-step explanation: