Which property allows you to write the expression

Based on the diagram shown, find θ to the nearest degree.

sattari [20]

Answer:

θ = 38°

Step-by-step explanation:

The lower right triangle is congruent to the upper left triangle, so we have θ and 20° being the two acute angles in the triangle. The law of sines tells you ...

sin(θ)/9 = sin(20°)/5

sin(θ) = (9/5)sin(20°)

θ = arcsin(9/5·sin(20°)) ≈ 38°

___

Another solution to the triangle is θ = 180° -38° = 142°. The diagram clearly shows θ as an acute angle, so we take this second solution to be extraneous.

8 0

3 years ago

(j-1)(-3) I need help solving the distributive property.

Reptile [31]

Answer:

Step-by-step explanation:

-3*j = -3j

(-3)(-1) = 3

Answer

-3j + 3 Notice the sign change. Be careful about that when signs are used with the distributive property.

5 0

3 years ago

Convert the unit<br>1.5kg to g

navik [9.2K]

Answer:

The answer should be 1500g I think.

6 0

3 years ago

Integer questions. Thank you!

andrezito [222]

3. We can see that
1+100=101
2+99=101
3+98=101
.................
and so on
From 1 to 100, there are 50 pairs that has the sum is 101
Z=101×50=5050
4. We can see that
1-2=-1
3-4=-1
5-6=-1
...........
and so on
There are 10 pairs that has sum is -1
Z= 10×(-1)=-10

4 0

3 years ago

Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunne

Pavlova-9 [17]

Answer:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

$p_v =P(t_{(15)}>1.507)=0.076$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X_{CTS}=2.35$ represent the mean for the sample CTS

$\bar X_{N}=1.83$ represent the mean for the sample Normal

$s_{CTS}=0.88$ represent the sample standard deviation for the sample of CTS

$s_{N}=0.54$ represent the sample standard deviation for the sample of Normal

$n_{CTS}=10$ sample size selected for the CTS

$n_{N}=7$ sample size selected for the Normal

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:

Null hypothesis: $\mu_{CTS} \leq \mu_{N}$

Alternative hypothesis: $\mu_{CTS} > \mu_{N}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{CTS}-\bar X_{N}}{\sqrt{\frac{s^2_{CTS}}{n_{CTS}}+\frac{s^2_{N}}{n_{N}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{CTS}+n_{N}-2=10+7-2=15$

Since is a one side right tailed test the p value would be:

$p_v =P(t_{(15)}>1.507)=0.076$

We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,TRUE)"

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

6 0

4 years ago