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Lera25 [3.4K]
3 years ago
15

Contact forces are actions that take place only when objects are touching each other. Non-contact forces are actions that can ta

ke place whether objects are touching or not. Which of the following is a non-contact force? A. impact force B. gravity C. air resistance D. pushing
Mathematics
1 answer:
Sindrei [870]3 years ago
4 0

Answer:

I am pretty sure it is Gravity..

Step-by-step explanation:

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Help will mark brainliest answer
Naddika [18.5K]

Answer : 110 degree

To find angle 1 , we apply outside angle theorem  Lets name each point  Measurement of arc EF=280 degrees  

Measurement of arc GH = 60

Angle D = angle 1

Please refer to the theorem attached below  

angle D = \frac{arc(EF)-arc(GH)}{2}

Now we plug in the values

angle 1 = \frac{280-60}{2}

angle 1 = 110

Measurement of angle 1 = 110 degrees

5 0
3 years ago
Assume that you have a sample of n 1 equals 6​, with the sample mean Upper X overbar 1 equals 50​, and a sample standard deviati
tigry1 [53]

Answer:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

df=6+5-2=9

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

n_1 =6 represent the sample size for group 1

n_2 =5 represent the sample size for group 2

\bar X_1 =50 represent the sample mean for the group 1

\bar X_2 =38 represent the sample mean for the group 2

s_1=7 represent the sample standard deviation for group 1

s_2=8 represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 > \mu_2

We are assuming that the population variances for each group are the same

\sigma^2_1 =\sigma^2_2 =\sigma^2

The statistic for this case is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

The pooled variance is:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

We can find the pooled variance:

S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67

And the pooled deviation is:

S_p=7.46

The statistic is given by:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

The degrees of freedom are given by:

df=6+5-2=9

The p value is given by:

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

4 0
3 years ago
I'm stuck- lol. I'll give brainliest to the first correct answer! :D
polet [3.4K]

Answer:

Its 4:)

Step-by-step explanation:

8 0
3 years ago
I still need help! Please! Help me
Svetach [21]

Part A = 60

Part B is C

6 0
3 years ago
Work out 12+8÷(9-5) 0.018÷0.06 Express as single fraction 5/7÷2/5
Anettt [7]

Step-by-step explanation:

I don't know if the first set of numbers is all in one set, but I'll do my best to give you an answer.

Really all you need to do is use PEMDAS for the first question.

(Parentheses, exponents, multiply, divide, add, subtract. In that order)

1 2 + 8 \div (9 - 5) \\ 12 + 8 \div 4 \\ 12 + 2 \\ 14

Then to simplify that fraction next to it, notice that 0.018 is 3x 0.06.

that's a 3:1 ratio, so it ends up simplifying to this:

\frac{3}{1}

Lastly, to solve the division of that fraction. If you divide by a fraction, you multiply whatever it's dividing by its inverse.

So...

\frac{5}{7}  \div  \frac{2}{5}  \\  \frac{5}{7}  \times  \frac{5}{2}  \\  \frac{25}{14}

7 0
3 years ago
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