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3 years ago

Evaluate the expression 3x - 2, for x = -10.

Mathematics

1 answer:

8090 [49]3 years ago

6 0

Answer:

-32

Step-by-step explanation:

substitute -10 for x and solve

3(-10)-2

3(-10) = -30

(-30) -2 =-32 ( same sign so you add two numbers and keep sign )

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The value of the definite integral / 212x – sin(x) dx / 122x-sin(x) is

blondinia [14]

Hi there!

$\boxed{= 70 + cos(12) - cos(2) \approx 71.26}$

$\int\limits^{12}_{2} {x-sin(x)} \, dx$

We can evaluate using the power rule and trig rules:

$\int x^n = \frac{x^{n+1}}{n+1}$

$\int x = \frac{1}{2}x^{2}$

$\int -sin(x) = cos(x)$

Therefore:

$\int\limits^{12}_{2} {x-sin(x)} \, dx = [\frac{1}{2}x^{2}+cos(x)]_{2}^{12}$

Evaluate:

$(\frac{1}{2}(12^{2})+cos(12)) - (\frac{1}{2}(2^2)+cos(2))\\= (72 + cos(12)) - (2 + cos(2))\\\\= 70 + cos(12) - cos(2) \approx 71.26$

3 0

3 years ago

Choose the equation below that represents the line passing through the point (1, -4) with a slope of 1/2

tatiyna

For this case we have that by definition, the line equation of the slope-intersection form is given by:

$y = mx + b$

Where:

m: It's the slope

b: It is the cut-off point with the y axis

According to the statement we have:

$m = \frac {1} {2}$

Thus, the equation is of the form:

$y = \frac {1} {2} x + b$

We substitute the given point and find "b":

$-4 = \frac {1} {2} (1) + b\\-4 = \frac {1} {2} + b\\-4- \frac {1} {2} = b\\b = - \frac {9} {2}$

Finally, the equation is of the form:

$y = \frac {1} {2} x- \frac {9} {2}$

Answer:

$y = \frac {1} {2} x- \frac {9} {2}$

6 0

3 years ago

Karina needs a total of $45 to buy her mother a birthday present. She has saved 20% of the amount so far. How much has she saved

professor190 [17]

Answer:

$9.00

Step-by-step explanation:

$45.00x20%=9

7 0

3 years ago

Read 2 more answers

Write the order of operations down using your own words

dybincka [34]

The order of operations is:

Parentheses, Exponent, Division, Multiplication, Addition, Subtraction.

Or PEMDAS.

hope it helps!

6 0

3 years ago

A survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries,

Andru [333]

Answer:

A. Null and alternative hypothesis:

$H_0: \mu_d=0\\\\H_a:\mu_d\neq 0$

B. Yes. At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

P-value = 0.00002

C. As the difference is calculated as (population 1 − population 2), being population 1: groceries and population 2: dinning out, and knowing there is evidence that the true mean difference is positive, we can say that the groceries annual credit card charge is higher than dinning out annual credit card charge.

The point estimate is the sample mean difference d=$840.

The 95% confidence interval for the mean difference between the population means is (490, 1190).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

Then, the null and alternative hypothesis are:

$H_0: \mu_d=0\\\\H_a:\mu_d\neq 0$

The significance level is 0.05.

The sample has a size n=42.

The sample mean is M=840.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1123.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{1123}{\sqrt{42}}=173.2827$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{840-0}{173.2827}=\dfrac{840}{173.2827}=4.848$

The degrees of freedom for this sample size are:

$df=n-1=42-1=41$

This test is a two-tailed test, with 41 degrees of freedom and t=4.848, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>4.848)=0.00002$

As the P-value (0.00002) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

We have to calculate a 95% confidence interval for the mean difference.

The t-value for a 95% confidence interval and 41 degrees of freedom is t=2.02.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=2.02 \cdot 173.283=350$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 840-350=490\\\\UL=M+t \cdot s_M = 840+350=1190$

The 95% confidence interval for the mean difference is (490, 1190).

7 0

2 years ago

Evaluate the expression 3x - 2, for x = -10.​

Evaluate the expression 3x - 2, for x = -10.