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stich3 [128]
4 years ago
8

20 POINTS! Name the property of equality you would use to solve 2 = 3/4 + x

Mathematics
1 answer:
N76 [4]4 years ago
3 0
Its eeither transposing or trabsposition
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Please help i dont remember how to do this
inessss [21]
Just add 5 5 and 10 10 to get yo answers
i will work
8 0
3 years ago
2.2y−(y−7.3)÷10=0.1<br><br>PLEASE HELP!<br><br>Answer:
Naddik [55]

Answer:

y = -0.525

Step-by-step explanation:

2.2y - y + 7.3 / 10 = 0.1

1.2y + 0.73 = 0.1

1.2y = -0.63

y = -0.525

6 0
3 years ago
Read 2 more answers
What is the probability of picking an orange marble and flipping tails?<br> ​
Anarel [89]

Answer:

1/7 and 1/2

Step-by-step explanation:

The probability of orange marble is 1/7

Probability of tails is 1/2

If you need to add this, then it'll be 1/14

7 x 2 = 14

1 x 1 = 1

7 0
2 years ago
The imaginary unit i is defined as i =
dangina [55]

i is defined as the square root of -1 or i=\sqrt{-1}

7 0
3 years ago
Read 2 more answers
Question 1 options:Residents in Portland, Oregon think that their city has more rainfall than Seattle, Washington. To test this
dimaraw [331]

Answer:

There is no enough evidence to to support the claim that Portland has more average yearly rainfall than Seattle.

Being μ1: average rainfall in Portland, μ2: average rainfall in Seattle, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 > 0

P-value = 0.1290

As the P-value is bigger than the significance level, the effect is not significant and the null hypothesis failed to be rejected.

Step-by-step explanation:

We have to test the hypothesis of the difference between means.

The claim is that Portland has more average yearly rainfall than Seattle.

Being μ1: average rainfall in Portland, μ2: average rainfall in Seattle, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 > 0

The significance level is 0.10.

The sample for Portland, of size n1=45, has a mean of M1=37.50 and standard deviation of s1=1.82.

The sample for Seattle, of size n1=35, has a mean of M1=37.07 and standard deviation of s1=1.68.

The difference between means is:

M_d= M_1-M_2=37.50-37.07=0.43

The standard error for the difference between means is:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1.82^2}{45}+\dfrac{1.68^2}{35}}=\sqrt{ 0.0736+0.0688 }=\sqrt{0.1424}\\\\\\s_{M_d}=0.3774

We can calculate the t-statistic as:

t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.43-0}{0.3774}=1.1393

The degrees of freedom are:

df=n1+n2-2=45+35-2=78

Then, the p-value for this one-tailed test with 78 degrees of freedom is:

P-value=P(t>1.1393)=0.1290

As the P-value is bigger than the significance level, the effect is not significant and the null hypothesis failed to be rejected.

There is no enough evidence to to support the claim that Portland has more average yearly rainfall than Seattle.

7 0
3 years ago
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