Ask question

Ask question

All categories

4 years ago

5

How do i answer this?? please help...

1 answer:

Kipish [7]4 years ago

7 0

The equation cannot be solved, but it can be simplified.

You can't add variables with different exponents. For example, you can't add

$x^2$ with $x^3$ because they have different exponents

You might be interested in

Use substitution to solve the system of equations.<br> x-2y=1<br> 3x-6y=3

Ivahew [28]

Change first equation
x - 2y = 1
add 2y to both side
x = 2y + 1
Plug into second equation
3(2y+1) - 6y = 3
Distributive property
6y + 3 - 6y = 3
y = 0, plug into first equation
x - 2(0) = 1, x = 1
Solution: x = 1, y = 0

7 0

3 years ago

How can a kepler-47c happen?? <br><br><br> Plz help :(

Natali5045456 [20]

Answer:

I'm not SUPER sure but, it only can exist where there is a region in a planetary system where liquid water might exist on the surface of a planet. That puts it in the "habitable zone" which is where it can exist.

6 0

3 years ago

xeze [42]

48/53

Steps

15 inches = 1.25 feet

Ratio = 12 : 1.25

Simplified = 9.6 : 1

9.6 + 1 = 10.6

9.6/10.6 = 48/53

Checking...

12 feet = 144 inches

Ratio = 144 : 15

144 + 15 = 159

144/159 = 48/53

5 0

3 years ago

Read 2 more answers

Preliminary data analyses indicate that it is reasonable to use a t-test to carry out the specified hypothesis test. Perform the

Readme [11.4K]

<h2>Answer with explanation:</h2>

By considering the given information, we have

Null hypothesis : $H_0: \mu=35.0$

Alternative hypothesis : $H_1: \mu\neq35.0$

Since the alternative hypothesis is two-tailed , so the test is a two-tailed test.

Given : Sample size : n= 20, since sample size is less than 30 so the test applied is a t-test.

$\overline{x}=41.0$ ; $\sigma= 3.7$

Test statistic : $t=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

i.e. $t=\dfrac{41.0-35.0}{\dfrac{3.7}{\sqrt{20}}}=7.252112359\approx7.25$

Degree of freedom : n-1 = 20-1=19

Significance level = 0.01

For two tailed, Significance level $=\dfrac{0.01}{2}=0.005$

By using the t-distribution table, the critical value of t = $t_{19, 0.005}=2.861$

Since , the observed t-value (7.25) is greater than the critical value (2.861) .

So we reject the null hypothesis, it means we have enough evidence to support the alternative hypothesis.

We conclude that there is some significance difference between the mean score for sober women and 35.0.

3 0

4 years ago

-5,15,-45,135 common ratio

S_A_V [24]

Answer:

5 is the common ratio for your numbers

7 0

3 years ago