1) Lynda runs the 400 meter for her school's track team. She beat her best time by 3.6

How do I solve <br>y=sin1/2(x+pi/4)

dexar [7]

Answer:

x = 2 csc(1) y - π/4

Step-by-step explanation:

Solve for x:

y = 1/2 sin(1) (x + π/4)

y = 1/2 (x + π/4) sin(1) is equivalent to 1/2 (x + π/4) sin(1) = y:

1/2 sin(1) (x + π/4) = y

Divide both sides by sin(1)/2:

x + π/4 = 2 csc(1) y

Subtract π/4 from both sides:

Answer: x = 2 csc(1) y - π/4

3 0

3 years ago

PLEASE HURRY ITS TIMED 20 POINTS!!!What is the y-coordinate of the point shown in the graph?

Drupady [299]

<h3><em>X CO-ORDINATE = 3</em></h3><h3><em>Y CO-ORDINATE = -7</em></h3><h3><em>HOPE IT HELPS.....BYE....</em></h3>

7 0

3 years ago

Read 2 more answers

Solve this equation <br><br> 1 . B/-7 = 4

weqwewe [10]

Step-by-step explanation:

B/-7 = 4

Cross multiple

-7×4 = B

-28 = B

8 0

3 years ago

Read 2 more answers

A corporate bond has a coupon rate of 5.5 percent, a $1,000 face value, and matures three years from today. The corporation is i

melomori [17]

Answer:

$= \frac{\frac{75}{100}\times 1000 + \frac{25}{100} \times \frac{60}{100}\times 1000 }{(1+\frac{15}{100})^3 }$

$=\frac{0.75\times 1000 + 0.25\times 0.60 \times 1000}{(1+0.15)^3}$

$=\frac{750+0.25\times 0.60\times 1000}{1.15^3} \\\\=\frac{750+150}{1.520875} =\frac{900}{1.520875} \\\\=591.76$

Step-by-step explanation:

= (probability of entire face value paid*face value+probability of entire face value not paid*percent of face value paid*face value)/(1+discount rate)^years to maturity

probability of entire face value paid = 75%

face value = 1000

probability of entire face value not paid = 25%

percent of face value paid= 60%

discount rate = 15%

years to maturity = 3

$= \frac{\frac{75}{100}\times 1000 + \frac{25}{100} \times \frac{60}{100}\times 1000 }{(1+\frac{15}{100})^3 }$

$=\frac{0.75\times 1000 + 0.25\times 0.60 \times 1000}{(1+0.15)^3}$

$=\frac{750+0.25\times 0.60\times 1000}{1.15^3} \\\\=\frac{750+150}{1.520875} =\frac{900}{1.520875} \\\\=591.76$

6 0

3 years ago

The degenerative disease osteoarthritis most frequently affects weight-bearing joints such as the knee. Two random samples from

Vinvika [58]

Answer:

$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$

$df=n_1 +n_2 -2=13+19-2=30$

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

$p_v =P(t_{30}>1.771)=0.0434$

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

Step-by-step explanation:

Data given

$\bar X_{1}=810$ represent the mean for sample 1

$\bar X_{2}=770$ represent the mean for sample 2

$s_{1}=67$ represent the sample standard deviation for 1

$s_{2}=56$ represent the sample standard deviation for 2

$n_{1}=13$ sample size for the group 2

$n_{2}=19$ sample size for the group 2

$\alpha=0.05$ Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for category 1 is higher than the mean for category 2, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=13+19-2=30$

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.

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

$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$

P value

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

$p_v =P(t_{30}>1.771)=0.0434$

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

4 0

3 years ago