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irga5000 [103]
3 years ago
14

A number is multiplied by 4 then 7 is added to the product after subtracting 3 the result is 8. what is the number​

Mathematics
2 answers:
ollegr [7]3 years ago
5 0

Answer:

1

Step-by-step explanation:

x*4+7-3=8

4x+7-3=8

4x+4=8

4x=4

x=1

soldi70 [24.7K]3 years ago
5 0

Answer:

the number is 1.

1 * 4 = 4

4 - 3 = 1

7 + 1 = 8

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Alice had 3 1/4 pounds of flour at the beginning of the week. At the end of the week, she had 2 3/4 pounds. How much has she use
Evgen [1.6K]

Answer:

1 2/4

Step-by-step explanation:

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4 0
3 years ago
A perpendicular bisector, , is drawn through point C on .
Liula [17]

Answer:

The x-intercept of CD is B(18/5,0). The point C(32,-71) lies on the line CD.

Step-by-step explanation:

the x-intercept of CD is[ A(3,0) B(18/5,0) C(9,0) D(45/2,0) ] . Point [ A(-52,117) B(-20,57) C(32,-71) D(-54,-128) ] lies on CD.

Given :

CD is perpendicular bisector of AB.

The coordinates of point A are (-3, 2) and the coordinates of point B are (7, 6).

C is the midpoint of AB.

C=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(\frac{7-3}{2},\frac{2+6}{2})=(2,4)

The coordinates of C are (2,4).

Line AB has a slope of:m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{6-2}{7-(-3)}=\frac{4}{10}=\frac{2}{5}

The product of slopes of two perpendicular lines is -1. Since the line CD is perpendicular to AB, therefore the slope of CD : m_2=-\frac{5}{2}

The point slope form of a line is given by:

y-y_1=m(x-x_1)

The slope of line CD is -\frac{5}{2} and the line passing through the point (2,4), the equation of line CD can be written as:

y-4=-\frac{5}{2}(x-2)\\y=-\frac{5}{2}x+5+4\\y=-\frac{5}{2}x+9       .... (1)

The equation of CD is y=-\frac{5}{2}x+9

In order to find the x-intercept, put y=0.

0=-\frac{5}{2}x+9\\\frac{5}{2}x=9\\x=\frac{18}{5}

Therefore the x-intercept of CD is B(18/5,0).

Put x=-52 in eq(1).

y=-\frac{5}{2}(-52)+9=139

Put x=-20 in eq(1).

y=-\frac{5}{2}(-20)+9=59

Put x=32 in eq(1)

y=-\frac{5}{2}(32)+9=-71

Put x=-54 in eq1).

y=-\frac{5}{2}(-54)+9=144

Thus, only point (32,-71) satisfies the equation of CD. Therefore the point C(32,-71) lies on the line CD.

8 0
3 years ago
What is the answer to 18×524
adoni [48]
Tha answer for the problem is 9,432
4 0
3 years ago
A student at a four-year college claims that mean enrollment at four-year colleges is higher than at two-year colleges in the Un
WINSTONCH [101]

Answer:

Part 1: The statistic

t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}} (1)  

And the degrees of freedom are given by df=n_1 +n_2 -2=35+35-2=68  

Replacing we got

t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31  

Part 2: P value  

Since is a right tailed test the p value would be:  

p_v =P(t_{68}>4.31)=0.000022 \approx 0.00002  

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

Step-by-step explanation:

Data given

\bar X_{1}=5135 represent the mean for four year college

\bar X_{2}=4436 represent the mean for two year college

s_{1}=783 represent the sample standard deviation for four year college

s_{2}=553 represent the sample standard deviation two year college

n_{1}=35 sample size for the group four year college

n_{2}=35 sample size for the group two year college

\alpha=0.01 Significance level provided

t would represent the statistic (variable of interest)  

System of hypothesis

We need to conduct a hypothesis in order to check if the mean enrollment at four-year colleges is higher than at two-year colleges in the United States , the system of hypothesis would be:  

Null hypothesis:\mu_{1}-\mu_{2}\leq 0  

Alternative hypothesis:\mu_{1} - \mu_{2}> 0  

We can assume that the normal distribution is assumed since we have a large sample size for each case n>30. So then the sample mean can be assumed as normally distributed.

Part 1: The statistic

t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}} (1)  

And the degrees of freedom are given by df=n_1 +n_2 -2=35+35-2=68  

Replacing we got

t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31  

Part 2: P value  

Since is a right tailed test the p value would be:  

p_v =P(t_{68}>4.31)=0.000022  

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

6 0
3 years ago
Once again I am in need of help,
natka813 [3]
Hope it helps...........


3 0
3 years ago
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