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2 years ago

What are the terms of each expression 3a-19

Mathematics

2 answers:

laiz [17]2 years ago

6 0

Hope this is what you were looking for:)

Rainbow [258]2 years ago

4 0

Answer:

The given expression has two terms "3a" and "19".

Step-by-step explanation:

The given expression is

$3a-19$

A term is either a single number or variable, or product of numbers and variables. The signs "+", "− " and "/" are used to separate the variables.

In the given expression the terms are separated through negative sign.

The above expression is the difference of two expression "3a" and "19".

Therefore the given expression has two terms "3a" and "19".

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2 years ago

Assume the blood pressures of 10 people were measured before and after sleeping for the night. What is the corresponding critica

NeX [460]

Answer:

For the critical value we need to calculate the degrees of freedom given by:

$df = n-1= 10-1=9$

And since we have a one tailed test we need to look in the t distribution with 9 degrees of freedom a quantile who accumulates 0.05 of the area on a tail and we got:

$|t_{crit}| =1.833$

Step-by-step explanation:

Previous concepts

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

Let put some notation

x=test value with right arm , y = test value with left arm

The system of hypothesis for this case are:

Null hypothesis: $\mu_y- \mu_x = 0$

Alternative hypothesis: $\mu_y -\mu_x \neq 0$

The first step is calculate the difference $d_i=y_i-x_i$

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{361}{5}$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}$

For the critical value we need to calculate the degrees of freedom given by:

$df = n-1= 10-1=9$

And since we have a one tailed test we need to look in the t distribution with 9 degrees of freedom a quantile who accumulates 0.05 of the area on a tail and we got:

$|t_{crit}| =1.833$

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2 years ago

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Thepotemich [5.8K]

Answer:

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Step-by-step explanation:

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2 years ago

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Margarita [4]

A)

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B)

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2 years ago

Conscientiousness is a tendency to show self-discipline, act dutifully, and aim for achievement. The trait shows a preference fo

MAVERICK [17]

Answer:

The Null hypothesis represented as

$H_0: \mu_1 - \mu_2 = 0$

The Alternative hypothesis represented as

$H_a: \mu_1 - \mu_2 < 0$

p-value = P(Z < -3.37 ) = 0.000376

There is insufficient evidence to conclude that graduate students score higher, on average, on the HPI than undergraduate students

Step-by-step explanation:

From the question we are told that

The population size is $n= 650$

The sample size for graduates is $n_1 = 300$

The sample mean for graduates is $\= x _1 = 148$

The sample standard deviation for graduates is $\sigma_1 = 16$

The sample size for under-graduates is $n _2 = 350$

The sample mean for under-graduates is $\= x _2 = 153$

The sample standard deviation for graduates is $\sigma_2 = 21$

The Null hypothesis represented as

$H_0: \mu_1 - \mu_2 = 0$

The Alternative hypothesis represented as

$H_a: \mu_1 - \mu_2 < 0$

Where $\mu_1 \ and \ \mu_2$ are the population mean

Now the test statistic is mathematically represented as

$t = \frac{(\= x_1 - \= x_2 ) }{ \sqrt{ \frac{ (n_1 - 1 )\sigma_1 ^2 + (n_2 - 1)\sigma_2^2}{n_1 +n_2 -2} } * \sqrt{ \frac{1}{n_1} + \frac{1}{n_2} } }$

substituting values

$t = \frac{(148 - 153 ) }{ \sqrt{ \frac{ (300- 1 )16 ^2 + (350 - 1) 21^2}{300 +350 -2} } * \sqrt{ \frac{1}{300} + \frac{1}{350} } }$

$t = -3.37$

The p-values is mathematically evaluated as

p-value = P(Z < -3.37 ) = 0.000376

The above answer is gotten using a p-value calculator at (0.05) level of significance

Looking the p-value we see that it is less than the level of significance (0.05) so Null hypothesis is rejected

Hence there is insufficient evidence to conclude that graduate students score higher, on average, on the HPI than undergraduate students

6 0

2 years ago