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4 years ago

6[13 – 2(4 + 1)] plz help mee

Mathematics

2 answers:

postnew [5]4 years ago

7 0

Answer:

18

Explanation:

Step 1 - Simplify the expression. To do this, add four and one.

6[13 - 2(4 + 1)]

6[13 - 2(5)]

6[13 - 2 × 5]

Step 2 - Simplify the expression. To do this, multiply two by five.

6[13 - 2 × 5]

6[13 - 10]

Step 3 - Simplify the expression. To do this, subtract ten from thirteen.

6[13 - 10]

6[3]

Step 4 - Simplify the expression. To do this, multiply six by three.

6[3]

Therefore, the answer to the given expression is 18.

Cloud [144]4 years ago

5 0

Answer:

18.

Step-by-step explanation:

6 * [13 - 2(4 + 1)] = 6 * [13 - 2(5)] = 6 * (13 - 10) = 6 * 3 = 18

Hope this helps!

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bekas [8.4K]

Answer:

$p_v =2*P(t_{149}>1.837)=0.0682$

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Step-by-step explanation:

Data given and notation

$\bar X=53000$ represent the sample mean

$s=20000$ represent the standard deviation for the sample

$n=150$ sample size

$\mu_o =50000$ 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 true mean is 50000, the system of hypothesis would be:

Null hypothesis: $\mu = 50000$

Alternative hypothesis: $\mu \neq 50000$

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:

$t=\frac{53000-50000}{\frac{20000}{\sqrt{150}}}=1.837$

Now we need to find the degrees of freedom for the t distirbution given by:

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What do you conclude? Use the p-value approach

Since is a two tailed test the p value would be:

$p_v =2*P(t_{149}>1.837)=0.0682$

So on this case the best answer would be:

b. Between 0.05 and 0.1

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4 years ago