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3 years ago

Please help me with this and i will reward 50 points if you answer all the areas.

2 answers:

7 0

C=total goals b=score started with 12 (idk, just say that you want to continue a game you’ve never finished before), ax=1 points per goal, and let's say c=winning score has to be greater than 50. x+12>50. SO for this, you could write that some people were playing a soccer match continuation. It was 9-0 in their last game, but they never finished, so they decided that they will continue from 9-0. Each score otherwise is 2 points. to win, you must get a greater score than 45
I hope this helps :3

NikAS [45]3 years ago

4 0

5x +3y>-15
5(0)+3(0)
0+0= 0 > -15

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The measure of one side of a square is (s + 3) inches. Write two different expressions to express the perimeter of this square.

valkas [14]

4(s+3)
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(s+3) + (s+3) + (s+3) + (s+3)

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3 years ago

Can sum1 answer these from least to greatest please?

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Step-by-step explanation:

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3 years ago

Suppose the expected tensile strength of type-A steel is 103 ksi and the standard deviation of tensile strength is 7 ksi. For ty

ExtremeBDS [4]

Answer:

i So the approximate distribution of $\= X$ is $\mu_{\= X} =103$ and $\sigma_{\= X} = 0.783$

ii So the approximate distribution of $\= Y$ is $\mu_{\= Y} =105$ and $\sigma_{\= Y} = 0.645$

the approximate distribution of $\=X - \= Y$ is $E (\= X - \= Y) = -2$ and $\sigma_{\= X - \=Y}=1.029$

Here we can see that the mean of the approximate distribution is negative which tell us that this negative value of the data for $\=X - \= Y$ sample are more and their frequency occurrence is higher than the positive values

the value of $P(-1 \le \=X - \= Y \le 1)$ is $= -0.1639$

Step-by-step explanation:

From the question we are given that

The expected tensile strength of the type A steel is $\mu_A = 103 ksi$

The standard deviation of type A steel is $\sigma_A = 7ksi$

The expected tensile strength of the type B steel is $\mu_B = 105\ ksi$

The standard deviation of type B steel is $\sigma_B = 5 \ ksi$

Also the assumptions are

Let $\= X$ be the sample average tensile strength of a random sample of 80 type-A specimens

Here $n_a =80$

Let $\= Y$ be the sample average tensile strength of a random sample of 60 type-B specimens.

Here $n_b = 60$

Let the sampling distribution of the mean be

$\mu _ {\= X} = \mu$

$=103$

Let the sampling distribution of the standard deviation be

$\sigma _{\= X} = \frac{\sigma }{\sqrt{n_a} }$

$= \frac{7}{\sqrt{80} }$

$=0.783$

So What this mean is that the approximate distribution of $\= X$ is $\mu_{\= X} =103$ and $\sigma_{\= X} = 0.783$

For $\= Y$

The sampling distribution of the sample mean is

$\mu_{\= Y} = \mu$

$= 105$

The sampling distribution of the standard deviation is

$\sigma _{\= Y} = \frac{\sigma }{\sqrt{n_b} }$

$= \frac{5}{\sqrt{60} }$

$= 0.645$

So What this mean is that the approximate distribution of $\= Y$ is $\mu_{\= Y} =105$ and $\sigma_{\= Y} = 0.645$

Now to obtain the approximate distribution for $\=X - \= Y$

$E (\= X - \= Y) = E (\= X) - E(\= Y)$

$= \mu_{\= X} - \mu_{\= Y}$

$= 103 -105$

$= -2$

The standard deviation of $\=X - \= Y$ is

$\sigma_{\= X - \=Y} = \sqrt{\sigma_{\= X}^2 - \sigma_{\= Y}^2}$

$= \sqrt{(0.783)^2 + (0.645)^2}$

$=1.029$

Now to find the value of $P(-1 \le \=X - \= Y \le 1)$

Let us assume that $F = \= X - \= Y$

$P(-1 \le F \le 1)$ $= P [\frac{-1 -E (F)}{\sigma_F} \le Z \le \frac{1-E(F)}{\sigma_F} ]$

$= P[\frac{-1-(-2)}{1.029} \le Z \le \frac{1-(-2)}{1.029} ]$

$= P[0.972 \le Z \le 2.95]$

$= P(Z \le 0.972) - P(Z \le 2.95)$

Using the z-table to obtain their z-score

$= 0.8345 - 0.9984$

$= -0.1639$

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3 years ago

For an analysis of variance comparing three treatment means, H0 states that all three population means are the same and H1 state

antoniya [11.8K]

Answer:

False

Step-by-step explanation:

The analysis of variance may be described as an hypothesis test which is used to make comparison between variables of two or more independent groups. The null hypothesis is always of the notion that there is no difference in the means. While the alternative hypothesis is the opposite, for two independent groups, the alternative hypothesis is that both means are different, or not equal or not the same. However. When we have more than 2 independent groups, then the alternative hypothesis is stated as : 'the means are not all equal'. This means that the means of each group does not all have to be different, but the mean of one group may be different from that of the other groups or the mean of two groups are different from the other groups and so on.

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3 years ago

Geometry First Semester.<br>Can someone please help with this?

Len [333]

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3 years ago