What is three equations that x equals 5

A random sample of 144 recent donations at a certain blood bank reveals that 81 were type A blood. Does this suggest that the ac

Reptile [31]

Answer:

Null hypothesis: $p=0.4$

Alternative hypothesis: $p \neq 0.4$

$z=\frac{0.5625 -0.4}{\sqrt{\frac{0.4(1-0.4)}{144}}}=3.98$

$p_v =2*P(z>3.98)=0.0000689$

Since the p value is very low compared to the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true percent of people with type A of blood is significantly different from 0.4 or 40%

Step-by-step explanation:

Information given

n=144 represent the random sample taken

X=81 represent the number of people with type A blood

$\hat p=\frac{81}{144}=0.5625$ estimated proportion of people with type A blood

$p_o=0.4$ is the value that we want to verify

$\alpha=0.01$ represent the significance level

z would represent the statistic

$p_v{/tex} represent the p value Hypothesis to testWe want to test if the percentage of the population having type A blood is different from 40%.: Null hypothesis:[tex]p=0.4$

Alternative hypothesis: $p \neq 0.4$

the statistic is given by:

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

Replacing the info given we got:

$z=\frac{0.5625 -0.4}{\sqrt{\frac{0.4(1-0.4)}{144}}}=3.98$

Now we can calculate the p value with this probability taking in count the alternative hypothesis:

$p_v =2*P(z>3.98)=0.0000689$

Since the p value is very low compared to the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true percent of people with type A of blood is significantly different from 0.4 or 40%

3 0

2 years ago

M1|L7 1 Decimal Round Round 5.489 to the nearest hundredth. ) Label the endpoints. )

vagabundo [1.1K]

Answer:

5.49

Step-by-step explanation:

5.489, the number 9 that is underlined is 5 or more so you put that at 0 then it would be 5.48, but since the 8 is also 5 or more and is the hundreths place you round that to 9 giving you 5.49

3 0

2 years ago

Suppose that you are responsible for making arrangements for a business convention and that you have been charged with choosing

REY [17]

Got a zero sum of what I was going to do get it back better than I did weewweee

8 0

2 years ago

What is the problem of this solving?!

nika2105 [10]

This question can be solved primarily by L'Hospital Rule and the Product Rule.

$y= \lim_{x \to 0} \frac{x^2cos(x)-sin^2(x)}{x^4}$

I) Product Rule and L'Hospital Rule:

$y= \lim_{x \to 0} \frac{[2xcos(x)-x^2sin(x)]-2sin(x)cos(x)}{4x^3}$

II) Product Rule and L'Hospital Rule:

$y= \lim_{x \to 0} \frac{[-2xsin(x)+2cos(x)]-[2xsin(x)+x^2cos(x)]-[2cos^2(x)-2sin^2(x)]}{12x^2} \\ y= \lim_{x \to 0} \frac{2cos(x)-4xsin(x)-x^2cos(x)-2cos^2(x)+2sin^2(x)}{12x^2}$

III) Product Rule and L'Hospital Rule:

$]y= \alpha + \beta \\ \\ \alpha =\lim_{x \to 0} \frac{-2sin(x)-[4sin(x)+4xcos(x)]-[2xcos(x)-x^2sin(x)]}{24x} \\ \beta = \lim_{x \to 0} \frac{4sin(x)cos(x)+4sin(x)cos(x)}{24x} \\ \\ y = \lim_{x \to 0} \frac{-6sin(x)-4xcos(x)-2xcos(x)+x^2sin(x)+8sin(x)cos(x)}{24x}$

IV) Product Rule and L'Hospital Rule:

$y = \phi + \varphi \\ \\ \phi = \lim_{x \to 0} \frac{-6cos(x)-[-4xsin(x)+4cos(x)]-[2cos(x)-2xsin(x)]}{24x} \\ \varphi = \lim_{x \to 0} \frac{[2xsin(x)+x^2cos(x)]+[8cos^2(x)-8sin(x)]}{24x}$

V) Using the Definition of Limit:

$y= \frac{-6*1-4*1-2*1+8*1^2}{24} \\ y= \frac{-4}{24} \\ \boxed {y= \frac{-1}{6} }$

3 0

2 years ago

Help quickly pls!!!

yan [13]

Answer:

A

Step-by-step explanation:

treat the problem as an equation and solve for n, it is 4

4 0

2 years ago