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pychu [463]
3 years ago
10

19/5 as a mixed number

Mathematics
2 answers:
Feliz [49]3 years ago
8 0
The answer would be 3 and 4 over 5
Reika [66]3 years ago
5 0
19 divided by 5 is 3 4/5
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Rory's battery was full at 6:00 a.M, or t = 0. Which times could Rory's phone have been plugged into the charger? Select three o
Taya2010 [7]

Complete question:

Rory records the percentage of battery life remaining on his phone throughout a day. The graph represents the percentage of battery life remaining after a certain number of hours.

Rory’s battery was full at 6:00 a.m, or t = 0. Which times could Rory's phone have been plugged into the charger? Select three options.

12:00 p.m.

3:00 p.m.

5:00 p.m.

8:00 p.m.

11:00 p.m.

The graph is attached

Answer:

3:00 p.m

5:00 p.m

11:00 p.m

Step-by-step explanation:

From the graph, there are rise times, fall times and there are times the movement is steady (no rise or fall).

The rise time represents the time the phone was plugged. The fall time represents when the charger has been unplugged so the battery level starts depreciating. When it is constant, it means the battery level is at 100 but the charger is still connected to the phone.

We are told at initial condition, time was 6 a.m (t = 0). To get the exact time, we are to add the initial condition, i.e 6

From the graph, the times the phone could have been plugged are 8:00 to 12:00 hrs and 16:00 to 20:00 hrs.

Converting from 24 hr to 12 hr time, we have:

6 + 8hrs = 14:00 = 2:00 p.m

6 + 9hrs = 15:00 = 3:00 p.m

6 + 10 hrs = 16:00 = 4:00 p.m

6 + 11 hrs = 17:00 = 5:00 p.m

6 + 12 hrs = 18:00 = 6:00 p.m

6 + 16 hrs = 22:00 = 10:00 p.m

6 + 17 hrs = 23:00 = 11:00 p.m

6 + 18 hrs = 00:00 = 12:00 am

6 + 19 = 1 : 00 = 1:00 a.m

6 + 20 = 2:00 = 2:00 a.m

From the options given in the question, we have:

3:00 p.m; 5:00 p.m; 11:00 p.m

Therefore, times could Rory's phone have been plugged into the charger are:

3:00 p.m

5:00 p.m

11:00 p.m

5 0
3 years ago
Solve the following system equation graphically on the set of axes below y= x+4 and y= -3/2 - 1 PLEASE HELP MEEE
gizmo_the_mogwai [7]

Answer:

y=4

y= -1

Step-by-step explanation:

8 0
2 years ago
Lucy is going to walk in a fundraising event to raise money for her school band. Her mother has agreed to donate $7.50 to the sc
Readme [11.4K]

Answer: y = $7.50x + $25

Step-by-step explanation:

From the question, we are informed that Lucy is going to walk in a fundraising event to raise money for her school band and her mother has agreed to donate $7.50 to the school for each mile that Lucy walks, plus an additional $25.

The equation that represents y, the total amount of money Lucy's mother will donate if Lucy walks x miles during the event will be:

= ($.750 × x) + $25

y = $7.50x + $25

Since Lucy walks x Mike's, we multiply x by $7.50 and then add the answer to the additional $25 which is the answer gotten above.

7 0
3 years ago
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
A red gumball is 2 5/8 Inches across. A green gumball is 2 5/6 inches across and a blue gumball in 2 7/9 inches across. list the
Masteriza [31]

Answer: 5 8 7 9 5 6

Step-by-step explanation:

Because you go by which is closer to the denominator

8 0
3 years ago
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