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monitta
3 years ago
11

Solve the following equations: (6x–5)–7(3x+10)=0

Mathematics
2 answers:
satela [25.4K]3 years ago
7 0

Answer:

-15

Step-by-step explanation:

6x-5-21x-70=0

-15x=0+5+70

-15x=75

x=75/-15

brilliants [131]3 years ago
3 0

Answer:

X= -5

Step-by-step explanation:

PEMDAS

(6x-5)-7(3x+10)=0

(6x-5)-21x-70=0

-15x-75=0

add 75 to 75 on the left add 75 to 0 on the right

-15x=75

divide both sides by -15

X= -5

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The answer to this question is negative and positive
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Select the sets of data where it would be better to use a histogram than a
Contact [7]

Answer:

Weight of students in your class

Daily rainfall in Miami over a month

Step-by-step explanation:

A histogram is applied when the data is discrete/ continuous to present a visual understanding of the data by showing the number of data set fitting is a specified range.

A dot plot utilizes data which has only one measured variable, where the variable can be organized into categories.

8 0
3 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
3 years ago
A serving of chickpeas contains 2,150 milligrams of potassium. How many grams of
Zepler [3.9K]

Answer: 109 mg

Step-by-step explanation:

5 0
3 years ago
See image attached below keeeeeeeeeeeeeeeeeeeed
emmasim [6.3K]

Answer:

.13%

68.26%

2.28%

47.72%

49.87%

34.13%

Step-by-step explanation:

1.) standardize by subtracting the mean and dividng by the standard deviation

(625-1000)/125= -3

go to a ztable to get .0013 or (1-.9987)

this is equal to .13%

2.)

875<x<1125

standardize both seperately and subtract them

(875-1000)/125= -1  whose probability is .1587 (or 1-.8413)

(1125-1000)/125= 1  whose probability is .8413

.8413-.1587= 68.26%

3.)

Find the probability that someone paid less than 1250 and take its compliment

(1250-1000)/125= 2 which has a probability of .9772

take its compliment (1-.9772)= .0228= 2.28%

4.)

Same process as question 2

(750-1000)/125= -2 which has a probability of (1-.9772) = .0228

(1000-1000)/125= 0 which has a probability of .5

.5-.0228= .4772= 47.72%

5.) same deal as the previous question

(625-1000)/125= -3 which has a probability of (1-.9987)= .0013

(1000-1000)/125= 0 which has a probability of .5

.5-.0013= .4987= 49.87%

6.)same deal the previous question

(875-1000)/125= -1 which has a probability of (1-.8413)=.1587

(1000-1000)/125= 0 which has a probability of .5

.5-.1587= .3413= 34.13%

4 0
3 years ago
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