Please help!!! ASAP the bottom number is 3 btw.

If cosх A/2 =3/5, find the value of cos A

nikitadnepr [17]

Answer:

cosA = - $\frac{7}{25}$

Step-by-step explanation:

Using the double angle identity

cos2A = 2cos²A - 1 , then

cosA = 2cos²( $\frac{A}{2}$ ) - 1 , that is

cosA = 2 × ( $\frac{3}{5}$ )² - 1

= 2 × $\frac{9}{25}$ - 1

= $\frac{18}{25}$ - $\frac{25}{25}$

= - $\frac{7}{25}$

5 0

2 years ago

8000 people attended a football

Lyrx [107]

Answer:

Step-by-step explanation:

Okay, so this is impossible to tell the amount of spaces, but let's say it was 4000 for VIP and 4000 for Popular Stands. Multiply $30 by 4000, and you get $120k Earnings. To check your answer, Divide 120,000 by 4000, and you get $30 for each stand.

Now, let's move on with the VIP stands. A VIP Stand costs $50, so 4,000 x $50 = $200k earnings. To check your answer, divide 200,000 by 4000, and you get $50 for each VIP Stand.

If you need to find each variant, go by 1k seat increments, i.e. 1000 and 7000, 2000 and 6000, etc.

Glad I could help!

-Vibilities

8 0

2 years ago

Which of the following numbers is not an integer? A -32 B 80 C 5.81 D 0 O<br>

larisa [96]

5.81

option C

hope it helps...!!!

3 0

2 years ago

Read 2 more answers

What is the equation of a line that passes through (-4, 3) and (6, -2) ?

s2008m [1.1K]

First find the slope
the slope of the line that passes through (x1,y1) and (x2,y2) is
(y2-y1)/(x2-x1)

given
(-4,3) and (6,-2)
the slope is (-2-3)/(6-(-4))=(-5)/(6+4)=-5/10=-1/2

so
the equation of a line tat passes through the point (x1,y1) and has a slope of m is
y-y1=m(x-x1)
use (-4,3)
and we found slope=-1/2
y-3=-1/2(x-(-4))
y-3=-1/2(x+4)
if yo want slope intercept form
y-3=-1/2x-2
y=-1/2x+1
if yo want standard form
1/2x+y=1
x+2y=2

8 0

3 years ago

A sample of 1200 computer chips revealed that 45% of the chips fail in the first 1000 hours of their use. The company's promotio

yaroslaw [1]

Answer:

$z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08$

$p_v = P(Z$

So the p value obtained was a low value and using the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of chips that fail in the first 1000 hours of their use is not significantly less than 0.48.

Step-by-step explanation:

Data given and notation

n=1200 represent the random sample taken

$\hat p=0.45$ estimated proportion of chips that fail in the first 1000 hours of their use

$\mu_0 =0.48$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion si less then 0.48:

Null hypothesis: $p\geq 0.48$

Alternative hypothesis: $p < 0.48$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.45 -0.48}{\sqrt{\frac{0.48(1-0.48)}{1200}}}=-2.08$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

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

$p_v = P(Z$

So the p value obtained was a low value and using the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of chips that fail in the first 1000 hours of their use is not significantly less than 0.48.

6 0

3 years ago