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Serga [27]
2 years ago
8

Peggy is thinking of a number such that when twice the number is added to three times one more than the number she gets the same

result as when she multiplies four times one less than the number. What number is Peggy thinking of?
Mathematics
1 answer:
mars1129 [50]2 years ago
4 0

Answer:

-7

Step-by-step explanation:

Lets assume the number Peggy is thinking of be "x".

Now as given, when twice the number is added to three times one more than the number. We can write it as

∴ 2x+3(x+1)--- Equation 1

Again, it is given that Peggy gets the same result as when she multiplies four times one less than the number.

∴ 4(x-1)-- equation 2

Next, we can equate both the equation 1 and 2 as it is given that result is same.

2x+3(x+1)= 4(x-1)

Let´s distribute 3 into (x+1) and 4 into (x-1)

⇒ 2x+3x+3=4x-4

⇒ 5x+3=4x-4

Subtract both side by 4x and 3

∴ x=-7

∴ Peggy was thinking of -7

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A phone manufacturer wants to compete in the touch screen phone market. Management understands that the leading product has a le
frutty [35]

Answer:

a) Null hypothesis:\mu_{1} - \mu_{2} \leq 120

Alternative hypothesis:\mu_{1} - \mu_{2}>120

b) Calculate the statistic

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

t=\frac{(485-340)-120}{\sqrt{\frac{55^2}{100}+\frac{30^2}{120}}}}=4.0689  

c) Critical value

The first step is calculate the degrees of freedom, on this case:

df=n_{1}+n_{2}-2=100+120-2=218

The significance level on this case is 0.05 or 5% so we need to find a quantile on the t distribution with 218 degrees of freedom that accumulates 0.95 of the area on the left and 0.05 of the area on the right tail and this value can be founded with the following excel code: "=T.INV(1-0.05,218)" and we got:

t_{crit}= 1.652

Conclusion

Since our calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 5% of significance and then we can conclude that the new product has a battery life more than two hours (120 minutes) longer than the leading product

Step-by-step explanation:

Data given and notation

\bar X_{1}= 8*60 +5 =485 represent the mean for the new sample

\bar X_{2}=5*60+40=340 represent the mean for the old sample

s_{1}=55 represent the sample standard deviation for the new sample

s_{2}=30 represent the sample standard deviation for the old sample

n_{1}=100 sample size selected for the new sample

n_{2}=120 sample size selected for the old sample

\alpha=0.05 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)

a) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the new product has a battery life more than two hours longer than the leading product., the system of hypothesis would be:

Null hypothesis:\mu_{1} - \mu_{2} \leq 120

Alternative hypothesis:\mu_{1} - \mu_{2}>120

If we analyze the size for the samples both are higher than 30 but we don't know the population deviations so for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{(\bar X_{1}-\bar X_{2})- \Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

b) Calculate the statistic

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

t=\frac{(485-340)-120}{\sqrt{\frac{55^2}{100}+\frac{30^2}{120}}}}=4.0689  

c) Critical value

The first step is calculate the degrees of freedom, on this case:

df=n_{1}+n_{2}-2=100+120-2=218

The significance level on this case is 0.05 or 5% so we need to find a quantile on the t distribution with 218 degrees of freedom that accumulates 0.95 of the area on the left and 0.05 of the area on the right tail and this value can be founded with the following excel code: "=T.INV(1-0.05,218)" and we got:

t_{crit}= 1.652

Conclusion

Since our calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 5% of significance and then we can conclude that the new product has a battery life more than two hours (120 minutes) longer than the leading product

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vekshin1

Answer:

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Step-by-step explanation:

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Answer:

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Answer:

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Step-by-step explanation:

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I have attached the work to your problem below.

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