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vlabodo [156]
3 years ago
9

Jason wants to mow his rectangular lawn. If his lawn is 15 feet by 25 feet how

Mathematics
1 answer:
Alika [10]3 years ago
6 0

Answer:

375

Step-by-step explanation:

To get the square feet of how many square feet he has to mow you have to multiply the length by the width which would be 15 x 25 to equal 375 square feet in total :)

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Can someone help with this question? Thank u! :)
MatroZZZ [7]

Answer:

2.50

Step-by-step explanation:

the pattern shown in the miles driven goes up each time by 5000, stating it's going up by a constant of 5000. the cost goes up each time by 2000.

divide 5000/2000 which would give you 2.5

round it by two decimal places which gives you 2.50

7 0
3 years ago
#3 is what I need help on
miv72 [106K]

Answer: 11.05

Step-by-step explanation:

Divide 5.20 by 8 to find the price of each bag (which is .65 cents.) Multiple .65 by 20 (13). 15% of 13 is 1.95. 13 - 1.95 is 11.05

PS: 1.95 is the discount. Hope this helps. Have a good day.

3 0
3 years ago
Read 2 more answers
Solve this answer<br> 5/3 + 2/5
salantis [7]
5/3+2/5=25/15+6/15=31/15=2 1/15

2 1/15 or 31/15

5 0
3 years ago
Read 2 more answers
Consider a chemical company that wishes to determine whether a new catalyst, catalyst XA-100, changes the mean hourly yield of i
kolezko [41]

Answer:

Null hypothesis:\mu = 750  

Alternative hypothesis:\mu \neq 750  

t=\frac{811-750}{\frac{19.647}{\sqrt{5}}}=6.943  

p_v =2*P(t_{4}>6.943)=0.00226  

If we compare the p value and a significance level assumed \alpha=0.05 we see that p_v so we can conclude that we reject the null hypothesis, and the actual true mean is significantly different from 750 pounds per hour.  

Step-by-step explanation:

Data given and notation

Data:    801, 814, 784, 836,820

We can calculate the sample mean and sample deviation with the following formulas:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=811 represent the sample mean  

s=19.647 represent the standard deviation for the sample

n=5 sample size  

\mu_o =750 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the mean is different from 750 pounds per hour, the system of hypothesis would be:  

Null hypothesis:\mu = 750  

Alternative hypothesis:\mu \neq 750  

Compute the test statistic  

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

We can replace in formula (1) the info given like this:  

t=\frac{811-750}{\frac{19.647}{\sqrt{5}}}=6.943  

Now we need to find the degrees of freedom for the t distirbution given by:

df=n-1=5-1=4

What do you conclude?  

Compute the p-value  

Since is a two tailed test the p value would be:  

p_v =2*P(t_{4}>6.943)=0.00226  

If we compare the p value and a significance level assumed \alpha=0.05 we see that p_v so we can conclude that we reject the null hypothesis, and the actual true mean is significantly different from 750 pounds per hour.  

4 0
3 years ago
The following are the last 10 run scores Colin got in cricket:
Softa [21]
A) mean score = (18+25+28+15+6+18+2+4+25+15)/10
= 15.6 runs
Approximately 16 runs
b) (18+25+28+15+6+18+2+4+25+15+31)/11
= 17 runs
8 0
3 years ago
Read 2 more answers
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