Can anybody help me plzzzz

help me on these 5 math problems, answer them correctly and explain your answers, please and thank you! <3 i have trouble on

BARSIC [14]

Answer: H

Step-by-step explanation: move the dilations

6 0

2 years ago

Read 2 more answers

Factor p^4-81 completely showing all work.

3241004551 [841]

Answer:

Step-by-step explanation:

$p^4-81=(p^2)^2-9^2=(p^2+9)(p^2-9)=(p^2-9(-1))(p+3)(p-3)\\=(p^2-9i^2)(p+3)(p-3)\\=(p+3i)(p-3i)(p+3)(p-3)$

8 0

2 years ago

1. (a). A police car, approaching right-angled intersection from the north, is chasing a speedmg

tamaranim1 [39]

Answer:

<em>The SUV is running at 70 km/h</em>

Step-by-step explanation:

<u>Speed As Rate Of Change </u>

The speed can be understood as the rate of change of the distance in time. When the distance increases with time, the speed is positive and vice-versa. The instantaneous rate of change of the distance allows us to find the speed as a function of time.

This is the situation. A police car is 0.6 Km above the intersection and is approaching it at 60 km/h. Since the distance is decreasing, this speed is negative. On the other side, the SUV is 0.8 km east of intersection running from the police. The distance is increasing, so the speed should be positive. The distance traveled by the police car (y) and the distance traveled by the SUV (x) form a right triangle whose hypotenuse is the distance between them (d). We have:

$d=\sqrt{x^2+y^2}$

To find the instant speeds, we need to compute the derivative of d respect to the time (t). Since d,x, and y depend on time, we apply the chain rule as follows:

$\displaystyle d\ '=\frac{x.x'+y.y'}{\sqrt{x^2+y^2}}$

Where x' is the speed of the SUV and y' is the speed of the police car (y'=-60 km/h)

We'll compute :

$d=\sqrt{(0.8)^2+(0.6)^2}=\sqrt{0.64+0.36}$

$d=1\ km$

We know d'=20 km/h, so we can solve for x' and find the speed of the SUV

$\displaystyle \frac{x.x'+y.y'}{\sqrt{x^2+y^2}}=20$

Thus we have

$x.x'+y.y'=20$

Solving for x'

$\displaystyle x'=\frac{20-y.y'}{x}$

Since y'=-60

$\displaystyle x'=\frac{20+0.6(60)}{0.8}$

$x'=70\ km/h$

The SUV is running at 70 km/h

7 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

3 years ago

Need Help Fast Please!!!!!

Arturiano [62]

Answer:

The fourth term in the sequence is equal to the third term plus the common difference, or 16+3=19- last choice

8 0

2 years ago