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rjkz [21]
3 years ago
11

What's the answer to this?

Mathematics
1 answer:
Dmitry [639]3 years ago
8 0

For problem 1, the number classification problem,  

Natural: 1

Whole: 0

Integer: -3

Rational: 6

Irrational: pi

When plotting pi, put it just in front of the number 3



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jayda bought a baseball hat for $8 when it was originally $18 dollars. what percent discount did she get?
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The percent discount was,

55.5556. It could be changed to 56%, but as a whole it would more than likely be 55%.

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It is 3 p.m. in the Central Time Zone; therefore, it is what time in the Mountain Time Zone.
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3 years ago
A school raised $4,111 in donations. They plan to use the money to buy new social studies textbooks. Each textbook costs $40. Es
Debora [2.8K]

Answer:

102 textbooks

Step-by-step explanation:

From the above question,

The total amount in donations = $4,111 . They plan to use the money to buy new social studies textbooks. Each textbook costs $40.

If :

$40 = 1 textbook

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Cross Multiply

40 × x textbooks = $4,111 × 1

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x = 102.775 textbooks

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4 0
3 years ago
Match each expression to the description shown. Answers may be used more than once.
Aliun [14]
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5 0
3 years ago
A sample of 41 observations is selected from one population with a population standard deviation of 3.4. The sample mean is 100.
Marizza181 [45]

Answer:

a) If we see the alternative hypothesis we see that we are conducting a bilateral test or two tail.

b) (-1.9886, 1.9886)

c) t=\frac{(100 -98.4)-(0)}{\sqrt{\frac{3.4^2}{41}}+\frac{5.6^2}{45}}=1.617

d) p_v =2*P(t_{84}>1.617) =0.1096

So with the p value obtained and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 is NOT significantly different than the mean for the group 2.  

e) p_v =2*P(t_{84}>1.617) =0.1096

We can use the following excel code to calculate it: "=2*(1-T.DIST(1.617;84;TRUE))"

Step-by-step explanation:

The system of hypothesis on this case are:

Null hypothesis: \mu_1 = \mu_2

Alternative hypothesis: \mu_1 \neq \mu_2

Or equivalently:

Null hypothesis: \mu_1 - \mu_2 = 0

Alternative hypothesis: \mu_1 -\mu_2\neq 0

Our notation on this case :

n_1 =41 represent the sample size for group 1

n_2 =45 represent the sample size for group 2

\bar X_1 =100 represent the sample mean for the group 1

\bar X_2 =98.4 represent the sample mean for the group 2

s_1=3.4 represent the sample standard deviation for group 1

s_2=5.6 represent the sample standard deviation for group 2

Part a

If we see the alternative hypothesis we see that we are conducting a bilateral test or two tail.

Part b

On this case since the significance level is 0.05 and we are conducting a bilateral test we have two critical values, and we need on each tail of the distribution \alpha/2 = 0.025 of the area.

The distribution on this cas since we don't know the population deviation for both samples is the t distribution with df=n_1+n_2 -2= 41+45-2=84 degrees of freedom.

We can use the following excel codes in order to find the critical values:

"=T.INV(0.025,84)", "=T.INV(1-0.025,84)"

And we got: (-1.9886, 1.9886)

Part c

The statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}

And now we can calculate the statistic:

t=\frac{(100 -98.4)-(0)}{\sqrt{\frac{3.4^2}{41}}+\frac{5.6^2}{45}}=1.617

The degrees of freedom are given by:

df=41+45-2=84

Part d

And now we can calculate the p value using the altenative hypothesis:

p_v =2*P(t_{84}>1.617) =0.1096

So with the p value obtained and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 is NOT significantly different than the mean for the group 2.  

Part e

p_v =2*P(t_{84}>1.617) =0.1096

We can use the following excel code to calculate it: "=2*(1-T.DIST(1.617;84;TRUE))"

8 0
3 years ago
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