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Archy [21]
3 years ago
8

3. What is the lowest common multiple (LCM) of 3 and 7?

Mathematics
1 answer:
jeka57 [31]3 years ago
3 0

Answer:

3= 3,6,9,12,15,18,21,24,27,30,33

7=7,14,21,28,.....

the least common multiple is 21

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Find the point that is 3/4 of the distance P (-6,-5) to the point Q (2,7)
Zarrin [17]

Using proportions, the coordinates of the point 3/4 of the way from P to Q are: (0,4).

<h3>What is a proportion?</h3>

A proportion is a fraction of a total amount, and the measures are related using a rule of three. Due to this, relations between variables, either direct or inverse proportional, can be built to find the desired measures in the problem.

We want to find the coordinates of point M(x,y) 3/4 of the way from P to Q, hence the rule is given by:

M - P = 3/4(Q - P)

For the x-coordinate, we have that:

x + 6 = 3/4(2 + 6)

x + 6 = 6

x  = 0.

For the y-coordinate, we have that:

y + 5 = 3/4(7 + 5)

y + 5 = 9.

y = 4.

The coordinates are (0,4).

More can be learned about proportions at brainly.com/question/24372153

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8 0
1 year ago
If p is inversely proportional to the square of q, and p is 25 when q is 3, determine p
ArbitrLikvidat [17]

Answer:

p = 9 when q = 5.

Step-by-step explanation:

p is inversely proportional to the square of q

This means that:

p = \frac{k}{q^2}

In which k is a constant multiplier.

p is 25 when q is 3

We use this to find k.

p = \frac{k}{q^2}

25 = \frac{k}{3^2}

k = 25*9 = 225

So

p = \frac{225}{q^2}

Determine p when q is equal to 5.

p = \frac{225}{q^2} = \frac{225}{5^2} = 9

p = 9 when q = 5.

6 0
2 years ago
A sheep is bought for £40 and sold for £33. What is the percentage loss?
valkas [14]

Answer:

£13  %11

Step-by-step explanation:

4 0
3 years ago
The mean time required to repair breakdowns of a certain copying machine is 93 minutes. The company which manufactures the machi
Sonja [21]

Answer:

t=\frac{88.8-93}{\frac{26.6}{\sqrt{73}}}=-1.349    

p_v =P(t_{(72)}  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, and we can't concluce that the true mean is less than 93 min at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=88.8 represent the sample mean

s=26.6 represent the sample standard deviation for the sample  

n=73 sample size  

\mu_o =93 represent the value that we want to test

\alpha=0.05 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.  

We need to conduct a hypothesis in order to check if the true mean i lower than 93 min, the system of hypothesis would be:  

Null hypothesis:\mu \geq 93  

Alternative hypothesis:\mu < 93  

If we analyze the size for the sample is > 30 but we don't know the population deviation so 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".  

Calculate the statistic

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

t=\frac{88.8-93}{\frac{26.6}{\sqrt{73}}}=-1.349    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=73-1=72  

Since is a one side test the p value would be:  

p_v =P(t_{(72)}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, and we can't concluce that the true mean is less than 93 min at 5% of signficance.  

5 0
3 years ago
Express the set x &gt; − 1 x&gt;-1 using interval notation
Ad libitum [116K]

x >  - 1 \\ ( - 1. \infty )
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