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gogolik [260]
3 years ago
7

How to find the LCM of 8 and 10

Mathematics
2 answers:
Katen [24]3 years ago
5 0

Answer:

El MCM entre 8 y 10 es: 40

Step-by-step explanation:

Es cuestión de simplificar ambos números:

8... 10| 2 (se puede sacar mitad)

4. 5| 2 (aún se puede sacar mitad)

2. 5| 2 (aún se puede sacar mitad)

1. 5| 5 (se puede sacar quinta)

1|

Se multiplican : 2*2*2*5: 40

(40 es múltiplo de 8 y de 10)

Espero te ayude

inn [45]3 years ago
4 0
8:8,16,24,32,40,48,56,64,72,80

10:10,20,30,40,50,60,70,80,90,100




Only 40 seems to be the lowest and common multiple 8 and 10 have

Hope this helps
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Romb ma obwód równy 32 cm. Jakie jest pole tego rombu, jeżeli jego wysokość ma 3 cm?
Goshia [24]

Answer:

24 cm²

Step-by-step explanation:

Z zadanego powyżej pytania uzyskano następujące dane:

Obwód (obwód) = 32 cm

Wysokość (h) = 3 cm

Obszar rombu (A) =?

Następnie określimy długość boku rombu. Można to uzyskać w następujący sposób:

Obwód, czyli obwód (P) = 32 cm

Strona (y) =?

P = 4 s

32 = 4 s

Podziel obie strony przez 4

s = 32/4

s = 8 cm

Stąd długość boku rombu wynosi 8 cm

Na koniec określimy obszar rombu w następujący sposób:

Długość boków = 8 cm

Wysokość (h) = 3 cm

Obszar rombu (A) =?

A = sh

A = 8 × 3

A = 24 cm²

Dlatego obszar rombu jest

24 cm².

8 0
3 years ago
Which expression is equivalent to 3(5+4)
guajiro [1.7K]

Answer:

3(15+12)

Step-by-step explanation:

3(5+4)

3( 5 x 3 =15 and 4 x 3= 12)

4 0
2 years ago
- 2x (a+h)^2+4(a+h)-12
hjlf
That’s the answer mama

7 0
3 years ago
given the following linear function sketch the graph of the function and find the domain and range. f(x)=-5x+4
velikii [3]

It's a linear function. We need only two points to sketch the graph.

f(x) = -5x + 4 → y = -5x + 4

for x = 0 → y = -5(0) + 4 = 0 + 4 = 4 → (0, 4)

for x = 1 → y= -5(1) + 4 = -5 + 4 = -1 → (1, -1)

The domain and the range is the set of all real numbers.

5 0
3 years ago
Consider the following data set. The data were actually collected in pairs, and each row represents a pair.
Anna11 [10]

Answer:

Step-by-step explanation:

Hello!

a.

Using the data sets you have to analyze them as if they are two independent samples.

The hypotheses are:

H₀: μ₁=μ₂

H₁: μ₁≠μ₂

α: 0.05

Assuming that both data sets are from a normal distribution and both population variances, although unknown, are equal, the statistic to use is:

t=\frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}

Sa^2= \frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2}

Sa^2= \frac{9*5.38+9*3.65}{10+10-2} =4.51

Sa= 2.12

t_{H_0}= \frac{(49.69-50.52)-0}{2.12*\sqrt{\frac{1}{10} +\frac{1}{10} } } = -0.875

The p-value for the two-tailed test is:

P(t_{18}≤-0.875) + P(t_{18}≥0.875)= P(t_{18}≤-0.875) + (1 - P(t_{18}≤0.875))= 0.1965 + ( 1 - 0.8035)= 0.393

Since there is no significant level I determined it at 5%, comparing it to the p-value, the hypothesis test is not significant. Meaning that the difference between the population means of both groups is equal to zero.

b.

Considering the given data as a paired sample, you have to determine the variable difference to conduct the test:

Xd: the difference between X₁ and X₂

Before the sample mean and sample standard deviation you have to calculate the difference between the observations of group 1 and group 2. (2nd attachment)

Mean X[bar]d= -0.83

Standard deviation Sd= 1.28

The hypotheses for the paired sample test are:

H₀: μd=0

H₁: μd≠0

α: 0.05

t= \frac{X[bar]d-Mud}{\frac{Sd}{\sqrt{n} } } ~~t_{n-1}

t_{H_0}= \frac{-0.83-0}{\frac{1.28}{ \sqrt{10} }} = -2.05

The p-value for the two-tailed test is:

P(t_{9}≤-2.05) + P(t_{9}≥2.05)= P(t_{9}≤-2.05) + (1 - P(t_{9}≤2.05))= 0.0353 + (1 - 0.9647)= 0.0706

Comparing the p-value with the significance level, the hypothesis test is not significant. Meaning that the population mean of the difference between group 1 and group 2 is equal to cero. There is no difference between the two groups.

c.

The value of the statistic in "a" is greater than the value of the statistic obtained in "b". Since there are two samples used an "a" the degrees of freedom of the test are the double as the ones used in "b". The p-value obtained in "b" is around half of the p-value of "a".

The main difference is that in "a" you compared two different samples but in "b" you compared two dependent samples when analyzing paired data the "effect of the individual" is removed from the equation.

I hope this helps!

3 0
3 years ago
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