Ask question

Ask question

All categories

3 years ago

6

Allen wants to buy a hover board that sells for $249.99 and a helmet that is $36.54. How much change will Allen receive if he gi

ves the cashier three $100.00 bills?

1 answer:

vlada-n [284]3 years ago

6 0

Answer:

$13.47

Step-by-step explanation:

i added the two numbers then divided by 300

You might be interested in

Dolbear's law states the relationship between the rate at which Snowy Tree Crickets chirp and the air temperature of their envi

erastovalidia [21]

25.21 this is corect :D

4 0

3 years ago

Need math help As soon as possible

ehidna [41]

i believe the answer is B

but i could always be wrong :(

8 0

3 years ago

Read 2 more answers

Which graph shows the solution to x-2y_>9

STatiana [176]

Answer:

U

Step-by-step explanation:

6 0

3 years ago

a company makes $10 000 in 2017. The projection of profit increase is fixed 40% from the previous year. Predict the company prof

cestrela7 [59]

Answer:

19600

Step-by-step explanation:

For every year the profit increases by 40% so to figure out the profit from 2017 to 2018 you do 10,000*1.4 to and get 14,000. To get 2019's profit you do the same process, but use 2018's profit of 14,000 instead of 2017's profit to get 19,600.

6 0

3 years ago

Tensile-strength tests were carried out on two different grades of wire rod. Grade 1 has 10 observations yielding a sample mean

LuckyWell [14K]

Answer:

$t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240$

$df = n_1 +n_2 -2 = 10+15-2= 23$

$p_v = 2*P(t_{23} >2.240) = 0.035$

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

Step-by-step explanation:

Data given

$\bar X_1 = 1085$ sample mean for group 1

$\bar X_2 = 1034$ sample mean for group 2

$n_1 = 10$ sample size for group 1

$n_2 = 15$ sample size for group 2

$s_1 = 52$ sample deviation for group 1

$s_2 = 61$ sample deviation for group 2

Solution

We want to check if the two means are equal so then the system of hypothesis are:

Null hypothesis: $\mu_1= \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

And the statistic is given by:

$t = \frac{\bar X_1 -\bar X_2}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}$

And replacing we got:

$t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240$

The degrees of freedom are given by:

$df = n_1 +n_2 -2 = 10+15-2= 23$

And the p value would be:

$p_v = 2*P(t_{23} >2.240) = 0.035$

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

8 0

3 years ago