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sladkih [1.3K]
3 years ago
9

How many 2-digit numbers are multiples of either 5 or 7?

Mathematics
2 answers:
Murljashka [212]3 years ago
5 0

Answer:

46

Step-by-step explanation:

Ktuvouyhouhyobuyoubyoubuybooybuouybuyobbouyouybouybobuyob

Svet_ta [14]3 years ago
4 0

Answer:

46

Step-by-step explanation:

he count of numbers from 1 to n that are divisible by k is integer part of (n/k).

So the numbers from 1 to 99 divisible by 3 is int(99/3) = 33

Those divisible by 5: int(99/5) = 19

From the sum of those two, subtract those divisible by 15 = int(99/15) = 6

That gives you 33+19-6 = 46.

Now subtract the 1 digit numbers: 3, 5, 6, 9, or four of them

46 - 4 = 42 is the answer.

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For Christmas, each member of a class
schepotkina [342]

Answer:

There are 32 pupils in the class

Step-by-step explanation:

Let's say there are N pupils in the class. Then each pupil must send N-1 cards - because it would make no sense to send one to themselves! So each of the N pupils send N-1 cards, which becomes 992 cards in total. In equation form, this is

N(N-1)=992\\N^2-N-992=0

This is a second degree polynomial, which has the solutions

N=\frac{-b\pm \sqrt{b^2-4\cdot a \cdot c}}{2a}

where a=1, b=-1, \text{and }c=-992

If we insert these numbers in the equation,

N=\frac{-(-1)\pm \sqrt{1^2-4*1*(-992)}}{2*1}\\ = \frac{1\pm \sqrt{1+4*992}}{2}\\= \frac{1 \pm 63}{2}

If we choose the solution with the minus sign, we get

N=-31

but this makes no sense! There can't be a negative number of pupils in the class!

So we choose the solution with the plus sign,

N=\frac{1+63}{2}\\ =\frac{64}{2}\\ =32

So there are 32 pupils in the class

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which equation represents a proportional relationship that has a constant of proportionality equal to ?
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If y=3(4+x) then how do you find x without needing to graph it
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Answer:

In the picture.

Step-by-step explanation:

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There were 2430 Major League Baseball games played in 2009, and the home team won the game in 53% of the games. If we consider t
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Answer:

z=\frac{0.53 -0.5}{\sqrt{\frac{0.5(1-0.5)}{2430}}}=2.958  

p_v =P(Z>2.958)=0.0015  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion is significantly higher than 0.5 .  

Step-by-step explanation:

1) Data given and notation  

n=2430 represent the random sample taken

\hat p=0.53 estimated proportion when the home team won the game

p_o=0.5 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is higher than 0.5 or 50%:  

Null hypothesis:p\leq 0.5  

Alternative hypothesis:p > 0.5  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.53 -0.5}{\sqrt{\frac{0.5(1-0.5)}{2430}}}=2.958  

4) Statistical decision  

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. The next step would be calculate the p value for this test.  

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

p_v =P(Z>2.958)=0.0015  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion is significantly higher than 0.5 .  

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