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2x − 4y = 40, −x + 2y = −20 Find x and y intercept

Basile [38]

Answer:

The x and y intercept are (20,0)

Step-by-step explanation:

2x-4y=40—eqn1

-x+2y=-20—eqn2

Make x the subject of eqn2

-x=-20-2y

x=20+2y

plug this into eqn 1

2(20+2y)+4y=40

40+4y+4y=40

8y=40-40

y=0

plug y=0 into eqn2

-x+2(0)=-20

-x=-20

x=20

8 0

2 years ago

1. Solve the system using substitution

Molodets [167]

X+y=8
y=3x

You would want to substitute the y for the 3x
x+3x=8

You can now add the x with the 3x to get 4x
4x=8

Now divide the 4 on both side to solve for x
x=2

Now to solve for y just substitute the x for the 2
y=3(2)

Now multiply the 3 and the 2 to solve for y.
y=6

So your corrct answer is B:(2,6)

6 0

3 years ago

9) A gardener has 1000 plants. He wants to plant these in such a way that the number of

tatyana61 [14]

Answer:

make sure you make me brianliest

3 0

3 years ago

3.4.2 Discuss: Relating to Functions helppp plzz gimme good exmpl ill give 100 points

goldenfox [79]

Answer:

the function f(x) = x + 1 adds 1 to any value you feed it. You give it a 5, this function will give you a 6: f(5) = 5 + 1 = 6. Functions do have a criterion they have to meet, though. And that is the x value, or the input, cannot be linked to more than one output or answer.

Step-by-step explanation:

we could define a function where the domain X is again the set of people but the codomain is a set of number. For example , let the codomain Y be the set of whole numbers and define the function c so that for any person x , the function output c(x) is the number of children of the person x.

hop u get it

7 0

3 years ago

A Statistics professor has observed that for several years students score an average of 114 points out of 150 on the semester ex

kirill [66]

Answer:

Reject $H_0$ . The change is statistically significant. The software does appear to improve exam scores.

Step-by-step explanation:

We are given that a Statistics professor has observed that for several years students score an average of 114 points out of 150 on the semester exam. The software is expensive, and the salesman offers to let the professor use it for a semester to see if the scores on the final exam increase significantly.

In the trial course that used this software, 218 students scored an average of 117 points on the final with a standard deviation of 8.7 points.

Let $\mu$ = mean scores on the final exam.

SO, Null Hypothesis, $H_0$ : $\mu \leq$ 114 points {means that the mean scores on the final exam does not increases after using software}

Alternate Hypothesis, $H_A$ : $\mu$ > 114 points {means that the mean scores on the final exam increase significantly after using software}

The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average points = 117 points

s = sample standard deviation = 8.7 points

n = sample of students = 218

So, test statistics = $\frac{117-114}{\frac{8.7}{\sqrt{218} } }$ ~ $t_2_1_7$

= 5.091

Now, P-value of the test statistics is given by the following formula;

P-value = P( $t_2_1_7$ > 5.091) = Less than 0.05%

Since, in the question we are not given with the level of significance at which hypothesis can be tested, so we assume it to be 5%. Now at 5% significance level, the t table gives critical value of 1.645 at 217 degree of freedom for right-tailed test. Since our test statistics is higher than the critical value of t as 5.091 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the change is statistically significant. The software does appear to improve exam scores.

3 0

4 years ago