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3 years ago

Answer these two questions.

Mathematics

1 answer:

Jlenok [28]3 years ago

4 0

Hey! Number one is x is less then 60 and number two is 2 is less then x, which is less then 8. Hope this helps:)

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1.What is the y-intercept of the plane whose equation is 4x + 5y – z = 20?

Katena32 [7]

In case of plane, for y intercept, x and z are o .

So we will get

4(0)+5y-0=20

5y=20

Dividing both sides by 5

y = 4

So the y intercept is (0,4,0).

In second question, x intercept is 1, y intercept is -1 and z intercept is 2 .

So the correct option is the third equation , that is x-y+2z=4 .

8 0

3 years ago

Read 2 more answers

A zoologist is studying four very closely related feline species. She wishes to compare their gestation periods. An observationa

diamong [38]

Answer:

$p_v= 0.01$

Since the significance level is 0.05 we see that $pv$ so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:

b. at least some, but not all, of the gestation periods across all four species are the same

Because is only to identify if AT LEAST one mean is different, NOT to conclude that the all the means are different.

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

Solution to the problem

The hypothesis for this case are:

Null hypothesis: $\mu_{A}=\mu_{B}=\mu_{C}= \mu_D$

Alternative hypothesis: Not all the means are equal $\mu_{i}\neq \mu_{j}, i,j=A,B,C,D$

In order to find the mean square between treatments (MSTR), we need to find first the sum of squares and the degrees of freedom.

If we assume that we have $p=4$ groups and on each group from $j=1,\dots,p$ we have $n_j$ individuals on each group we can define the following formulas of variation:

$SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2$

$SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2$

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2$

And we have this property

$SST=SS_{between}+SS_{within}$

And in order to test this hypothesis we need to ue an F statistic and for this case the p value calculated is

$p_v= 0.01$

Since the significance level is 0.05 we see that $pv$ so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:

b. at least some, but not all, of the gestation periods across all four species are the same

Because is only to identify if AT LEAST one mean is different NOT to conclude that the all the means are different.

8 0

3 years ago

Jack bought a pair of pants and a shirt for $56. If the price of a shirt is two-fifth of the price of a pair of pants, then what

Juliette [100K]

Answer:

Fraction: 22 2/5

Decimal: $22.4

Step-by-step explanation:

2/5 x 56

6 0