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katen-ka-za [31]
2 years ago
9

Which number line represents the solution set for the inequality –negative StartFraction one-half EndFraction x is greater than

or equal to 4.x ≥ 4? A number line from negative 10 to 10 in increments of 2. A point is at negative 2 and a bold line starts at negative 2 and is pointing to the left. A number line from negative 10 to 10 in increments of 2. A point is at negative 8 and a bold line starts at negative 8 and is pointing to the left. A number line from negative 10 to 10 in increments of 2. A point is at negative 2 and a bold line starts at negative 2 and is pointing to the right. A number line from negative 10 to 10 in increments of 2. A point is at negative 8 and a bold line starts at negative 8 and is pointing to the right.

Mathematics
2 answers:
lozanna [386]2 years ago
4 0

Answer:

<h2>Second choice.</h2>

Step-by-step explanation:

The given inequality is

-\frac{1}{2}x \geq  4

Let's solve for x

x\leq -4(2)\\x\leq -8

Basically, the solution of the given inquality is set with all real numbers which are equal or less than -8. So, the solution must indicate a blue line starting at -8 pointing to its left.

Therefore, the second choice represents the solution to the given inequality.

dangina [55]2 years ago
4 0

Answer:

B

Step-by-step explanation:

You might be interested in
The coefficient of determination: a. cannot be negative b. is the square root of the coefficient of correlation c. can be negati
patriot [66]

Answer:

Correct statement: (a) and (b).

Step-by-step explanation:

R-squared is a statistical quantity that measures, just how near the values are to the fitted regression line. It is also known as the coefficient of determination.

The coefficient of determination is the squared value of the correlation coefficient.

That is,

R^{2}=(r(X, Y))^{2}

The correlation coefficient between two variables is the measure of the strength and direction of the relationship between two variables.

Its values ranges from -1.00 to +1.00.

Although <em>r</em> can be negative but R² is always positive, since square of a negative number is also positive.

So the correct statements are:

a. cannot be negative

b. is the square root of the coefficient of correlation

5 0
2 years ago
Which equation can be used to solve for b? Triangle A B C is shown. Angle B C A is a right angle and angle C A B is 30 degrees.
FinnZ [79.3K]

Answer: We should use the correct formula for tan α  (which is opposite

leg over adjacent leg)   tan α = 5/b

Step-by-step explanation:

See attached file

We must use the for equation of tan α = BC/AC  

tan α = BC/ b              tan α = 5/b      tan 30⁰  = 1/√3

b = 5/√3    ⇒ b = 5/1.7320

b = 2.8868  cm

4 0
2 years ago
Read 2 more answers
Triangle H N K is shown. Angle H N K is 90 degrees. The length of hypotenuse H K is n, the length of H N is 12, and the length o
Mazyrski [523]

Answer:

13

Step-by-step explanation:

From the question, we are given a triangle HNK with an angle of 90°

The length of hypotenuse H K is n,

the length of HN is 12

the length of N K is 6.

From the above values, obtained in the question, we can see that this is a right angled triangle.

We are asked to find the length of the hypotenuse.

We can use Pythagoras Theorem of solve for this.

c² = a² + b²

where c = HK = n

a = NK = 6

b = HN = 12

c² = 6² + 12²

c² = 36 + 144

c² = 180

c = √180

c = 13.416407865

Approximately to the nearest whole number = 13

Therefore the value of HK = n = 13

We can also use Law of Cosines as given in the question to solve for this.

a² = b² + c² - 2ac × Cos A

where c = HK = n

a = NK = 6

b = HN = 12

Hence

c² = a² + b² - 2ab × Cos C

c = √ (a² + b² - 2ab × Cos C)

Where C = 90

c = √ 6² + 12² - 2 × 6 × 12 × Cos 90

c = 13.42

Approximately to the nearest whole number ≈ 13

Therefore the value of HK = n = 13

8 0
1 year ago
A control chart is developed to monitor the analysis of iron levels in human blood. The lines on the control chart were obtained
sergejj [24]

Answer:

Step-by-step explanation:

Hello!

The variable is X: iron level on human blood.

It has a mean of μ= 51.50 mg/dl and a standard deviation of σ= 3.50 mg/dl.

According to the control chart, the process should be shut down for troubleshooting when the analysis shows values X[bar]≥ 53.42 mg/dl and X[bar]≤ 49.58 mg/dl

Warnings are received at levels X[bar]≥52.78 mg/dl and X[bar]≤ 50.22 mg/dl

The system works between levels 49.58<X[bar]<53.42 and works without warnings between 50.22<X[bar]<52.78.

Using these parameters you have to analyze if the lists of sample means to see which ones are within the working values are wich ones are outside this interval.

<u>Sample 1</u>

Min= 50.15

Max= 51.99

Mean= 51.61

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

<em>This sample's min value is below the lower limit of the warning interval but not low enough to reach action levels, the max value is within the working range.</em>

<em><u /></em>

<u>Sample 2 </u>

Min= 50.32

Max= 52.56

Mean= 51.16

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

<em>Both the max and min values of the sample are within the working range without warning.</em>

<em />

<u>Sample 3</u>

Min= 50.25

Max= 53.12

Mean= 51.83

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

<em>The max value of this sample is above the upper limit of the warning interval but does not surpass the upper bond of the troubleshoot interval. Min value is within working values.</em>

<em />

<u>Sample 4</u>

Min= 50.05

Max= 53.01

Mean= 51.70

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

<em>Both min and max values surpass the warning interval but do not reach action levels</em>.

<u>Sample 5</u>

Min= 50.35

Max= 52.71

Mean= 51.37

Without warning 50.22<X[bar]<52.78

Without Action interval 49.58<X[bar]<53.42

<em>Both min and max values are within working levels without warnings.</em>

<em />

Considering that samples reaching warning levels should be shut down as a precaution, they are classified as:

Shutdown: Sample 1, 3 and 4

Do not shutdown: Sample 2 and 5.

I hope it helps!

8 0
2 years ago
Most US adults have social ties with a large number of people, including friends, family, co-workers, and other acquaintances. I
xxTIMURxx [149]

Answer:

t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

p_v =2*P(t_{(1699)}>1.971)=0.049  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

Step-by-step explanation:

Data given and notation  

\bar X=669 represent the sample mean

s=732 represent the sample standard deviation

n=1700 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0. represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is different from 634, the system of hypothesis would be:  

Null hypothesis:\mu = 634  

Alternative hypothesis:\mu \neq 634  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=1700-1=1699  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(1699)}>1.971)=0.049  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

3 0
1 year ago
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