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Natasha2012 [34]
3 years ago
15

(−7, 0), y = −0.3x + 4.3

Mathematics
1 answer:
Maurinko [17]3 years ago
6 0
Parlleral line? answer is y=0
You might be interested in
What is the value of x in the number 5/3+ 3x+4=8​
amm1812
5/3+3x+4= 8
= 3x= 4-5/3
= 3x= 7/3
= x= 7/9
4 0
3 years ago
Leon wants to estimate the proportion of the seniors at his high school who like to watch football. He interviews a simple rando
cupoosta [38]

Answer:

Yes, the random conditions are met

Step-by-step explanation:

From the question, np^ = 32 and n(1 − p^) = 18.

Thus, we can say that:Yes, the random condition for finding confidence intervals is met because the values of np^ and n(1 − p^) are greater than 10.

Also, Yes, the random condition for finding confidence intervals is met because the sample size is greater than 30.

Confidence interval approach is valid if;

1) sample is a simple random sample

2) sample size is sufficiently large, which means that it includes at least 10 successes and 10 failures. In general a sample size of 30 is considered sufficient.

These two conditions are met by the sample described in the question.

So, Yes, the random conditions are met.

5 0
3 years ago
Solve x2-8x=3 by completing the square. which is the solution set of the equation
Mazyrski [523]

we have

x^{2}-8x=3

Complete the square. Remember to balance the equation by adding the same constants to each side

x^{2}-8x+16=3+16

x^{2}-8x+16=19

Rewrite as perfect squares

(x-4)^{2}=19

(x-4)^{2}=19\\ (x-4)=(+/-)\sqrt{19} \\ \\ x1=4+\sqrt{19} =8.359\\ x2=4-\sqrt{19} =-0.359

therefore

the answer is

x1=4+\sqrt{19} =8.359\\x2=4-\sqrt{19} =-0.359

5 0
3 years ago
Read 2 more answers
What % of 800,000 = 560,000​
Ne4ueva [31]

Answer:

70%

Step-by-step explanation:

560,000/800,000= 0.70

6 0
3 years ago
If y-3x=9, 7x+y=25, what is the value of x and y?
lianna [129]

\bold{\rm{Given}}\text{ : If y - 3x = 9, 7x + y = 25, what is the value of x and y}?

\bold{\rm{Solution}} \text{: We'll solve this by substitution method}

\dashrightarrow  y - 3x = 9 \\  \\  \dashrightarrow   \boxed{y = 9 + 3x } \qquad .. \sf eq(1)

\text{we got the value of y now getting the value of x}

\looparrowright 7x + y = 25 \\  \\  \looparrowright 7x = 25 - y \\  \\  \looparrowright  \boxed{x =   \frac{25 - y}{7}} \qquad .. \sf eq(2)

\text{Now, putting y = 9 + 3x in eq(2)}

\hookrightarrow  x =  \frac{25 - y}{7}   \\  \\ \hookrightarrow  x =  \frac{25 - 9 + 3x}{7}  \\  \\ \hookrightarrow  7x =  25 - 9 + 3x \\  \\ \hookrightarrow  7x - 3x =  16 \\  \\ \hookrightarrow  4x = 16 \\  \\ \hookrightarrow  x =  \frac{16}{4}  \\  \\ \star \quad  \boxed{ \green{ \frak{ x = 4}}}

\text{Now we know the value of x, for getting y we need to put x in eq(1)}

: \implies y = 9 + 3x \\  \\   : \implies y = 9 + 3(4) \\  \\  : \implies y = 9 + 12  \\  \\    \star \quad  \boxed{\frak{  \green{y = 21}}}

\text{Hence, values of x and y are} \:  \frak{ \green{4}}  \: \text{and} \:  \frak{ \green{21}} \: \text{respectively}.

4 0
2 years ago
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