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Crazy boy [7]
3 years ago
13

Please help with 6-8 ! will give brainliest !! :)

Mathematics
1 answer:
Sauron [17]3 years ago
3 0

Answer:

6:6=x

7:16=y

8:20=z

Step-by-step explanation:

For 6, since BD is a perpendicuar bisector, that means 15x=90. Divide 90 by 15 and you get 6

for 7, if bd is a angle bisector, you can set 3y=48. divide 48 by 3 and you get 16

for 8, since Bd is a median, you can set 2z-3=z+17. Isolate the variable and you will get 2z-z=17+3. Add like terms together and you get z=20

Just so you know, a perpendicular bisector will make both angles equal to 90, an angle bisector make both angles equal to each other, and medians make it so that it hit a line in the center therefor you can set both sides of the line to be equal

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1. How can you find answers to the statistical questions?
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Make sure the sample size is big enough to model differences with a normal population. ...

Find the mean of the difference in sample proportions: E(p1 - p2) = P1 - P2 = 0.52 - 0.47 = 0.05.

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Find the probability.

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3 years ago
Tensile-strength tests were carried out on two different grades of wire rod. Grade 1 has 10 observations yielding a sample mean
earnstyle [38]

Answer:

t=\frac{(1085 -1034)-(0)}{57.646\sqrt{\frac{1}{10}+\frac{1}{15}}}=2.167

p_v =2*P(t_{23}

We see that the p value is lower than the significance level provided of 0.05 so then we have enough evidence to conclude that the true means are different at 5% of significance.

Step-by-step explanation:

Data given

n_1 =10 represent the sample size for group 1

n_2 =15 represent the sample size for group 2

\bar X_1 =1085 represent the sample mean for the group 1

\bar X_2 =1034 represent the sample mean for the group 2

s_1=52 represent the sample standard deviation for group 1

s_2=61 represent the sample standard deviation for group 2

We are assuming that the two samples are normally distributed with equal variances and that means:

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

System of hypothesis

Null hypothesis: \mu_1 = \mu_2

Alternative hypothesis: \mu_1 \neq \mu_2

The statistic 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 degrees of freedom are given by:

n_1+n_2 -2

And 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}

Replacing we got:

S^2_p =\frac{(10-1)(52)^2 +(15 -1)(61)^2}{10 +15 -2}=3323.043

And the deviation would be:

S_p=57.646

The degrees of freedom are:

df=10+15-2=23

The statistic would be:

t=\frac{(1085 -1034)-(0)}{57.646\sqrt{\frac{1}{10}+\frac{1}{15}}}=2.167

The p value would be

p_v =2*P(t_{23}

We see that the p value is lower than the significance level provided of 0.05 so then we have enough evidence to conclude that the true means are different at 5% of significance.

6 0
3 years ago
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