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3 years ago

Find the least common multiple of 10x4 and 6w3 .

1 answer:

8 0

10x⁴ and 6w³

1) We see no the same letters in both expressions, so we will need to include all letters in the final answer. x⁴w³

2) number part
10 = 2*5
6=3*2
We need 2*3*5=30 (we need 2 only one times, because we have 2 only one time in each of the numbers).
3)The least common multiple
2*3*5x⁴w³ = 30x⁴w³
Answer is 30x⁴w³.

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Fiona was thinking of a number. Fiona doubles it, then adds 5 to get an answer of 49.2. What was the original number?

Musya8 [376]

22.1
She added 5 so you take away 5 which would be 44.2 then she doubles it another word for multiplication so you do the opposite and divide by 2 and you get 22.1

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Wanna see me run to that hill?<br><br> Wanna see me do it again?

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Answer: uhh no but thanks for the points

Step-by-step explanation:

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3 years ago

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Jacob Lee is a frequent traveler between Los Angeles and San Diego. For the past month, he wrote down the flight times in minute

valkas [14]

Solution :

We know that

$H_0: \mu_1 = \mu_2=\mu_3$

$H_1 :$ At least one mean is different form the others (claim)

We need to find the critical values.

We know k = 3 , N = 35, α = 0.05

d.f.N = k - 1

= 3 - 1 = 2

d.f.D = N - k

= 35 - 3 = 32

SO the critical value is 3.295

The mean and the variance of each sample :

Goust Jet red Cloudtran

$\overline X_1 =50.5$ $\overline X_2 =50.07143$ $\overline X_3 =55.71429$

$s_1^2=19.96154$ $s_2^2=14.68681$ $s_3^2=36.57143$

The grand mean or the overall mean is(GM) :

$\overline X_{GM}=\frac{\sum \overline X}{N}$

$=\frac{51+51+...+49+49}{35}$

= 52.1714

The variance between the groups

$s_B^2=\frac{\sum n_i\left( \overline X_i - \overline X_{GM}\right)^2}{k-1}$

$=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$

$=\frac{127.1143}{2}$

= 63.55714

The Variance within the groups

$s_W^2=\frac{\sum(n_i-1)s_i^2}{\sum(n_i-1)}$

$=\frac{(14-1)19.96154+(14-1)14.68681+(7-1)36.57143}{(14-1)+(14-1)+(7-1)}$

$=\frac{669.8571}{32}$

= 20.93304

The F-test statistics value is :

$F=\frac{s_B^2}{s_W^2}$

$=\frac{63.55714}{20.93304}$

= 3.036212

Now since the 3.036 < 3.295, we do not reject the null hypothesis.

So there is no sufficient evidence to support the claim that there is a difference among the means.

The ANOVA table is :

Source Sum of squares d.f Mean square F

Between 127.1143 2 63.55714 3.036212

Within 669.8571 32 20.93304

Total 796.9714 34

3 0

3 years ago

What’s the vertical asymptote of the equation f(x)=log3(x-5)

ICE Princess25 [194]

Answer:The line x=5

Step-by-step explanation:

Asymptote of a function means the straight line closest to some part of the function which tends to ∞ or -∞. We know that ln x or log x has asymptote x=0. Here , the function is f(x) = log 3(x-5), so, the vertical asymptote will be the line x =5.

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3 years ago

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alina1380 [7]

Answer:

One

Step-by-step explanation:

x+4y = 6

if why is 2x-3 plug it in

x+4(2x-3) = 6

simplify it and you get

x=2

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