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2 years ago

Six farmers each have 6 barrels. In

Mathematics

2 answers:

AveGali [126]2 years ago

6 0

Answer:

I believe it's 60 legs.

Step-by-step explanation:

Each farmer has 2 legs so 2x6=12

Each cat has 4 legs so 4x6=24

Each kitten has 4 legs so 4x6=24

When you add all the sums up, you get 60.

zysi [14]2 years ago

3 0

Answer:

There would be 876 legs.

Step-by-step explanation:

To do it you could first find how many legs there are on one farm. SO you would take the 6 cats in one barrel. Then times it by the 6 kittens each of them have. That's the number of cats. Then times than number by 4 to get the total number of legs in one barrel. Then you add in the two legs of the farmer. Then times that total number by 6 to get all the legs on all six farms.

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Solve each system by elimination.

strojnjashka [21]

Answer:

y = -1, x = -2

Step-by-step explanation:

First, multiply the top equation by 2 to make 3x become 6x:

6x - 8y = -4

Second, multiply the bottom equation by 3 to make -2x become -6x:

9y - 6x = 3

Add the two equations together, and you get:

y = -1

Plug in y as -1 into the first or second equation:

3x - 4(-1) = -2

3(-1) - 2x = 1

x = -2

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3 years ago

How many times greater is one hour than one minute

OverLord2011 [107]

60. 1 minute is 1 and 60 is 60. Therefor an hour is 60 times more than 1.

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3 years ago

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The interior angles of a hexagon are in the ratio 1:2:3:4:5:9 Find the angles. This is an example of a concave hexagon. Write in

algol [13]

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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

2 years ago

Which choice below is a common multiple of 8 and 10 (but not the least common multiple).

Ilya [14]

Answer:

160

Step-by-step explanation:

4 is too low, 90 cant be divided by 8 and 160 is the only answer even though it is highest number

8 0

3 years ago

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