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2 years ago

14

Rudy, Maria, and Tito have a total of x textbooks Rudy has 6 textbooks andTito and Maria each have y textbooks. How many textboo

ks does Tito have interms of x ?

1 answer:

shtirl [24]2 years ago

6 0

Answer:

Tito have $\frac{x-6}{2}$ textbooks in terms of x

Step-by-step explanation:

Rudy has 6 textbooks

We are given that Tito and Maria each have y textbooks

So, Tito has no. of books = y

Maria has no. of books = y

So, total number of textbooks = 6+y+y=6+2y

We are given that Rudy, Maria, and Tito have a total of x textbooks

So,6+2y=x

So, 2y=x-6

$y=\frac{x-6}{2}$

Hence Tito have $\frac{x-6}{2}$ textbooks in terms of x

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The sum of two consecutive even integers divided by four is 189.5

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The sum of two consecutive even integers is a+(a+2) and divided by four is
(a+(a+2))/4 = 189.5
(2a+2)/4 = 189.5
2a+2 = 189.5 * 4
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2a = 758 - 2
2a = 756
a = 756/2 = 378
first number is a = 378
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1 year ago

Kaitlyn gave 1/2 of her candy bar to arianna. arianna gave 1/3 of the candy she got from kaitlyn to cameron. what fraction of a

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Let the total candy bar be represented by x.

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= $\frac{1}{3}* \frac{1}{2} x = \frac{1}{6} x$

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2 years ago

The table below shows the different times it takes Paula to drive to work each day depending on the rate at which she drives.

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Speed = distance / time
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2 years ago

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Step-by-step explanation:

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2 years ago

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We're testing the hypothesis that the average boy walks at 18 months of age (H0: p = 18). We assume that the ages at which boys

marusya05 [52]

Answer:

II. This finding is significant for a two-tailed test at .01.

III. This finding is significant for a one-tailed test at .01.

d. II and III only

Step-by-step explanation:

1) Data given and notation

$\bar X=19.2$ represent the battery life sample mean

$\sigma=2.5$ represent the population standard deviation

$n=25$ sample size

$\mu_o =18$ 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)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean battery life is equal to 18 or not for parta I and II:

Null hypothesis: $\mu = 18$

Alternative hypothesis: $\mu \neq 18$

And for part III we have a one tailed test with the following hypothesis:

Null hypothesis: $\mu \leq 18$

Alternative hypothesis: $\mu > 18$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-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".

3) Calculate the statistic

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

$z=\frac{19.2-18}{\frac{2.5}{\sqrt{25}}}=2.4$

4) P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=25-1=24$

Since is a two tailed test for parts I and II, the p value would be:

$p_v =2*P(t_{(24)}>2.4)=0.0245$

And for part III since we have a one right tailed test the p value is:

$p_v =P(t_{(24)}>2.4)=0.0122$

5) Conclusion

I. This finding is significant for a two-tailed test at .05.

Since the $p_v$ . We reject the null hypothesis so we don't have a significant result. FALSE

II. This finding is significant for a two-tailed test at .01.

Since the $p_v >\alpha$ . We FAIL to reject the null hypothesis so we have a significant result. TRUE.

III. This finding is significant for a one-tailed test at .01.

Since the $p_v >\alpha$ . We FAIL to reject the null hypothesis so we have a significant result. TRUE.

So then the correct options is:

d. II and III only

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2 years ago