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marishachu [46]

3 years ago

6

If △ABC is an isosceles triangle where AB¯≅AC¯, m∠A=(2x−20)°, and m∠B=(3x+5)°, then m∠C=__________°.

2 answers:

natta225 [31]3 years ago

6 0

M<c=11 thats the answwer

kotegsom [21]3 years ago

3 0

<span><span> Let ABC be an isosceles triangle with AB = AC. Let M denote the midpoint of BC (i.e., M is the point on BC for which MB = MC). Thena)<span> </span>Triangle ABM is congruent to triangle ACM.b)<span> </span>Angle ABC = Angle ACB (base angles are equal)c)<span> </span>Angle AMB = Angle AMC = right angle.d)<span> </span>Angle BAM = angle CAM

</span></span>

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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

Are O, N, and P collinear? If so, name the line on which they lie.

san4es73 [151]

We are given a figure where M, N , O and P points are given.

We need to explain if points O, N, and P collinear or not.

Note: All co-linear points are in a straight line.

From the figure, we can see that <em>O and N points are in a straight line but point P is aside.</em>

So, the points O, N, and P are not collinear.

Therefore, correct option is "<u>No, the three points are not collinear</u>".

7 0

3 years ago

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tiny-mole [99]

(x² + 16)(x² - 16)
= (x²)² - 16²
= x⁴ - 256

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coldgirl [10]

Answer:

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48 + 90 + y = 180

138 + y = 180

-138 + 138 + y = 180 - 138

y = 42°

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rusak2 [61]

6 is the constant in that equation. The answer is 6 because it is the only number that is not attached to a variable.

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3 years ago

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