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andre [41]
3 years ago
8

A bag contains 18 coins consisting of quarters and dimes. The total value of the coins is $2.85. Which system of equations can b

e used to determine the number of quarters, q, and the number of dimes, d, in the bag?
Mathematics
2 answers:
tatuchka [14]3 years ago
5 0
There's nothing here to choose from but it should be something like : q + d = 18; .25q + .10d = 2.85
mixas84 [53]3 years ago
3 0

Answer:

Simultaneous Equation

Step-by-step explanation:

A bag contains 18 coins consisting of quarters and dimes. The total value of the coins is $2.85. Which system of equations can be used to determine the number of quarters, q, and the number of dimes, d, in the bag?

To get the number of dimes and the number of quarters q will definitely have to be by simultaneous equation

let the number of dimes be d

let the number of quarters be q

let the cost of quarters/ one  be Q

let the cost of dime/one be  D

q+d=18--------------1

Qq+Dd=2.85.........2

from equation 1

q=18-d

substituting the value of q into equation 2

Q(18-d)+Dd=2.85

if cost of quarters/ one  is given and the cost of dime/one is also given we can go ahead to find

q and d

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Daily-high temperature measurements for 40 consecutive days are recorded for a particular city. The mean daily-high temperature
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Answer:

t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108    

p_v =P(t_{39}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.  

(D) P-val=0.021, fail to reject the null hypothesis

Step-by-step explanation:

1) Data given and notation  

\bar X=21.5 represent the mean for the temperatures

s=1.5 represent the sample standard deviation

n=40 sample size  

\mu_o =22 represent the value that we want to test

\alpha=0.01 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.  

We need to conduct a hypothesis in order to check if the mean is less than 22C, the system of hypothesis would be:  

Null hypothesis:\mu \geq 22  

Alternative hypothesis:\mu < 22  

If we analyze the size for the sample is > 30 and we don't know the population deviation so 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".  

Calculate the statistic

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

t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108    

P-value

We can calculate the degrees of freedom like this:

df=n-1=40-1=39

Since is a one left tailed test the p value would be:  

p_v =P(t_{39}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.  

The best option would be:

(D) P-val=0.021, fail to reject the null hypothesis

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Answer:

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Step-by-step explanation:

I am not sure if this is what you were looking for but it might be useful.

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