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Vladimir79 [104]
3 years ago
6

A recipe for a cassetole calls for 3 4/7 cups of rice.adam has already put in 3 1/4 cups how mqny more cups does he need to put

in
Mathematics
1 answer:
Maru [420]3 years ago
3 0
You need to make the fractions into improper fractions with the same denometer. 3 4/7 to an improper fraction is 25/7 and 3 1/4 is 13/4. Now get common denometers, in going to choose 28. 100/28 is the same as 25/7 and 91/28 is the same as 13/4. Now subtract 100/28 and 91/28 and you get 9/28. So you need to add 9/28 of a cup.
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Is 0.734734734 rational or irrational
Zepler [3.9K]

Answer: Rational

Step-by-step explanation:

Repeating decimals can be written as fractions, so they are rational.

6 0
3 years ago
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Someone help I think this is probability but I don’t know
pickupchik [31]

ANSWER

1. 3391680

2. 3876

EXPLANATION

1. We want to find the value of:

P^{12}_7

This is called "12 permutation 7"

Permutation function is defined as follows:

P^n_r\text{ = }\frac{n!}{(n-r)!}

Therefore, we have that:

\begin{gathered} P^{12}_7=\frac{12!}{(12\text{ - 7)}!}\text{ = }\frac{12!}{5!} \\ P^{12}_7=\frac{12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5!}{5!} \\ P^{12}_7=\text{ 12 }\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6 \\ P^{12}_7=3391680 \end{gathered}

2. We want to find:

C^{19}_{15}

This is called "19 combination 15"

Combination function is defined as:

C^n_r=\frac{n!}{(n-r)!\cdot r!_{}}

Therefore, we have that:

\begin{gathered} C^{19}_{15}\text{ = }\frac{19!}{(19-15)!\cdot15!}\text{ = }\frac{19!}{4!\cdot15!} \\ ^{}C^{19}_{15}=\frac{19\cdot18\cdot17\cdot16\cdot15!}{4\cdot3\cdot2\cdot1\cdot15!} \\ C^{19}_{15}=\frac{19\cdot18\cdot17\cdot16}{4\cdot3\cdot2\cdot1} \\ C^{19}_{15}=\text{ 3876} \end{gathered}

7 0
1 year ago
A researcher expects a treatment to produce an increase in the population mean. The treatment is evaluated using a one tailed hy
Shkiper50 [21]

Answer:

A) The researcher should reject the null hypothesis with a = .05 but not with a = .01.

Step-by-step explanation:

State the null and alternative hypotheses to be tested  

On this case we can assume that the researcher conduct a one right tailed test hypothesis like on this way:  

Null hypothesis:\mu \leq \mu_o  

Alternative hypothesis:\mu > \mu_o

The statistic for this case 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".  

On this case we have the value for z score given z =+1.85

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05,0.01. The next step would be calculate the p value for this test.  

Since is a one right tail test the p value would be:  

p_v =P(z>1.85)=1-P(Z  

If we compare the p value obtained and the significance level given \alpha=0.05,0.01 we can take a decision:

A) The researcher should reject the null hypothesis with a = .05 but not with a = .01.

YES, since the p value it's higher than the significance level of 0.01 so we FAIL to reject the null hypothesis at this significance level. And for the significance level 0.05 we have that p_v and we can reject the null hypothesis.

B) The researcher should reject the null hypothesis with either a = .05 or a = .01.

No since the p value it's higher than the significance level of 0.01 so we FAIL to reject the null hypothesis at this significance level

C) The researcher should fail to reject H0 with either a = .05 or a = .01.

No, for the significance level of 0.05 we reject the null hypothesis since p_v

D) This cannot be answered without additional information.

No, we have enough info to answer the question

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Bracelet

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