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3 years ago

If it rains, then the game will be canceled Hypothesis: Conclusion:

Mathematics

1 answer:

yarga [219]3 years ago

6 0

<h3>Answers:</h3>

Hypothesis = It rains

Conclusion = The game is canceled

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Explanation:

The general format for a conditional is "If P, then Q".

P is the hypothesis and Q is the conclusion. They are placeholders.

In this case

P = It rains
Q = The game is canceled

Some textbooks will write $P \to Q$ to indicate "if P, then Q"

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Ally's little sister wanted to help her make some oatmeal cookies. first ,she put 5/8 cup 0f oatmeal in the bowl. next,she added

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15/8 cups or 1 and 7/8 of acup

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4 years ago

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PLEASE HELP. I’ll mark you as brainliest if correct

kodGreya [7K]

Answer:

Both of them are D

Step-by-step explanation:

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3 years ago

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N is a positive integer

Murrr4er [49]

Part (1)

n is some positive integer. Let's say for now that n is even. So n = 2k, for some integer k

This means n-1 = 2k-1 is odd since subtracting 1 from an even number leads to an odd number.

Now multiply n with n-1 to get

n(n-1) = 2k(2k-1) = 2m

where m = k(2k-1) is an integer

The result 2m is even showing that n(n-1) is even

------------

Let's say that n is odd this time. That means n = 2k+1 for some integer k

And also n-1 = 2k+1-1 = 2k showing n-1 is even

Now multiply n and n-1

n(n-1) = (2k+1)(2k) = 2k(2k+1) = 2m

where m = k(2k+1) is an integer

We've shown that n(n-1) is even here as well.

------------

So overall, n(n-1) is even regardless if n is even or if n is odd.

Either n or n-1 will be even. If you multiply an even number with any number, the result will be even.

=======================================================

Part (2)

n is some positive integer

2n is always even since 2 is a factor of 2n

2n+1 is always odd because we're adding 1 to an even number. The sequence of integers goes even,odd,even,odd, etc and it does this forever.

-----------

Another way to see how 2n+1 is odd is to divide 2n+1 over 2 and you'll find that we get (2n+1)/2 = 2n/2+1/2 = n+0.5

The 0.5 at the end is not an integer, so there's no way that (2n+1)/2 is an integer; therefore 2n+1 is odd.

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3 years ago

Iron deficiency anemia is an important nutritional health problem in the U.S. A dietary assessment was performed on 51 boys 9-11

Bond [772]

Answer:

a) Null hypothesis: $\mu = 14.44$

Alternative hypothesis: $\mu \neq 14.44$

b) $t=\frac{12.50-14.44}{\frac{4.75}{\sqrt{51}}}=-2.917$

The degrees of freedom are given by:

$df =n-1= 51-1=50$

Now we can calculate the p value taking in count the alternative hypothesis

$p_v =2*P(t_{50}$

Since the p value is lower than the significance $\alpha=0.05$ we have enough evidence to reject the null hypothesis and the true mean is significantly different from 14.44

c) $12.50 -2.01 \frac{4.75}{\sqrt{51}}= 11.163$

$12.50 +2.01 \frac{4.75}{\sqrt{51}}= 13.837$

Step-by-step explanation:

Information given

$\bar X=12.50$ represent the mean for the daily iron intake

$s=4.75$ represent the sample deviation

$n=51$ sample size

$\mu_o =14.44$ represent the reference value

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic

$p_v$ represent the p value for the test

Part a

We want to test if the mean iron intake among the low-income group is different from that of the general population, the system of hypothesis would be:

Null hypothesis: $\mu = 14.44$

Alternative hypothesis: $\mu \neq 14.44$

Part b

Since we don't know the population deviation the statistic would be given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info we got

$t=\frac{12.50-14.44}{\frac{4.75}{\sqrt{51}}}=-2.917$

The degrees of freedom are given by:

$df =n-1= 51-1=50$

Now we can calculate the p value taking in count the alternative hypothesis

$p_v =2*P(t_{50}$

Since the p value is lower than the significance $\alpha=0.05$ we have enough evidence to reject the null hypothesis and the true mean is significantly different from 14.44

Part c

The confidence interval would be given by:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$

For the 95% confidence interval we can find the critical value in a t distribution with 50 degrees of freedom and we got:

$t_{\alpha/2}= 2.01$

And replacing we got:

$12.50 -2.01 \frac{4.75}{\sqrt{51}}= 11.163$

$12.50 +2.01 \frac{4.75}{\sqrt{51}}= 13.837$

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4 years ago

Arrange the given polynomial in descending order.<br> 9x? + 9x +3+2x

WARRIOR [948]

20x+3
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3 years ago