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Andrews [41]
3 years ago
6

Solve the system of equations using substitution.

Mathematics
1 answer:
Dafna11 [192]3 years ago
8 0
It’s B
2(-4)+8(4)=24
-8+32=24

-2(-4)+4=12
8+4=12
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In a test (+4) marks are given for every correct answers and (-3) marks are given for every incorrect answers 1) Tahiya answered
stealth61 [152]

Answer:

Tahiya attempted 3 incorrect answers.

Helmin attempted 11 incorrect answers.

Step-by-step explanation:

Correct = +4

Incorrect = -3

1) Tahiya got 12 correct answers and scored a 40, the number of incorrect answers was:

4*12-3I=40\\I=\frac{8}{3}\\I=2.67

2) Helmin got 5 correct answers and scored a -14, the number of incorrect answers was:

4*5-3I=-14\\I=\frac{34}{3}\\I=11.33

In both cases the number of incorrect answers obtained is a fraction, which means that either the data provided is inaccurate, or that the scores obtained by the students are approximate. Let's round the number of incorrect answers to the nearest whole answer.

Tahiya attempted 3 incorrect answers.

Helmin attempted 11 incorrect answers.

4 0
3 years ago
I just started the problem, but I'm a bit lost on what to do now.
borishaifa [10]
According to Greg himself (probably a master at pies) says perfect cherry pies have a ratio of 240 cherries to 3 pies. First, we need to divide 240 cherries by 3 pies to find out how many cherries are in one pie.

240 / 3 = 80

Then multiply by 9 to get your answer.

9 * 80 = 720
8 0
3 years ago
The breaking strengths of cables produced by a certain manufacturer have a mean, u, of 1750 pounds, and a standard deviation of
AlladinOne [14]

Answer:

We accept the null hypothesis that the  breaking strength mean is less and equal to 1750 pounds and has not increased.

Step-by-step explanation:

The null and alternative hypotheses are stated as

H0:  u ≥ 1750   i.e the mean is less and equal to 1750

against the claim

Ha: u > 1750  ( one tailed test)  the mean is greater than 1750

Sample mean = x`= 1754

Population mean = u = 1750

Population deviation= σ = 65 pounds

Sample size= n = 100

Applying the Z test

z= x`- u / σ/ √n

z= 1754- 1750 / 65/ √100

z= 4/6.5

z= 0.6154

The significance level alpha = 0.1

The z - value at 0.1 for one tailed test is ± 1.28

The critical value is z > z∝.

so

0.6154 is < 1.28

We accept the null hypothesis that the  breaking strength mean is less and equal to 1750 pounds and has not increased.

8 0
3 years ago
Explanation pls? step by step i get confused on these questions! thank you!​
eduard

Answer:

5xyz

Step-by-step explanation:

10x^2 y^2z and 15xyz^2

Rewriting

2*5*x*x*y*y*z   and 3*5 *x*y*z*z

What is common to both terms

5 x y z

The greatest common factor is 5xyz

8 0
3 years ago
Read 2 more answers
A. Find the linear approximating polynomial for the following function centered at the given point a. b. Find the quadratic appr
Tema [17]

Answer:

a. p1(x) = 2 - x

b. p2(x) = x² - 3*x + 3

c. p1(0.97) = 1.03; p2(0.97) = 1.0309

Step-by-step explanation:

f(x) = 1/x

f'(x) =  -1/x²

f''(x) = 2/x³

a = 1

a. The linear approximating polynomial is:

p1(x) = f(a) + f'(a)*(x - a)

p1(x) = 1/1 + -1/1² * (x - 1)

p1(x) = 1 - x + 1

p1(x) = 2 - x

b. The quadratic approximating polynomial is:

p2(x) = p1(x) + 1/2 * f''(a)*(x - a)²

p2(x) = 2 - x + 1/2 * 2/1³ * (x - 1)²

p2(x) = 2 - x + (x - 1)²

p2(x) = 2 - x + x² - 2*x + 1

p2(x) = x² - 3*x + 3

c. approximate 1/0.97 using p1(x)

p1(0.97) = 2 - 0.97 = 1.03

approximate 1/0.97 using p2(x)

p2(0.97) = 0.97² - 3*0.97 + 3 = 1.0309

7 0
3 years ago
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