All categories

3 years ago

Whats the perfect square of 63 rounding it to the hundreths place

Mathematics

1 answer:

Leni [432]3 years ago

4 0

The answer to your question is 7.94 that the perfect square root of 63

You might be interested in

Please Help, Geometry find exact value of missing sides

zzz [600]

Answer:

answer bottom is 90

Step-by-step explanation:

i helps

3 0

3 years ago

3/5÷33/5 find the quotient

defon

Flip 33/5

3/5 times 5/33
3 times 5 /5 times 33
15/165
Reduce
1/11

7 0

3 years ago

An auto mechanic ordered eight car spark plugs and four truck spark plugs. Car spark plugs cost $.75, and truck spark plugs cost

Sidana [21]

$.75(8) + $3.00(4) =
$6 + $12 = $18

7 0

3 years ago

Find a function where f(0)=2 and f(1)=2

xenn [34]

Answer:

Do you want to be extremely boring?

Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?

$f(x) = 2$ is a valid solution.

Want something more fun? Why not a parabola? $f(x)= ax^2+bx+c$ .

At this point you have three parameters to play with, and from the fact that $f(0)=2$ we can already fix one of them, in particular $c=2$ . At this point I would recommend picking an easy value for one of the two, let's say $a= 1$ (or even $a=-1$ , it will just flip everything upside down) and find out b accordingly: $f(1)=2 \rightarrow 1^2+b+2=2 \rightarrow b=-1$

Our function becomes

$f(x) = x^2-x+2$

Notice that it works even by switching sign in the first two terms: $f(x) = -x^2+x+2$

Want something even more creative? Try playing with a cosine tweaking it's amplitude and frequency so that it's period goes to 1 and it's amplitude gets to 2: $f(x) = A cos (kx)$

Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need $A= 2$ , and at that point the first condition is guaranteed; using the second to find k we get $2= 2 cos (k1) = cos k = 1 \rightarrow k = 2\pi$

$f(x) = 2cos(2\pi x)$

Or how about a sine wave that oscillates around 2? with a similar reasoning you get

$f(x)= 2+sin(2\pi x)$

Sky is the limit.

8 0

2 years ago

According to a polling organization, 24% of adults in a large region consider themselves to be liberal. A survey asked 200 resp

Varvara68 [4.7K]

Answer:

$z=\frac{0.375 -0.24}{\sqrt{\frac{0.24(1-0.24)}{200}}}=4.47$

Now we can calculate the p value based on the alternative hypothesis with this probability:

$p_v =P(z>4.47)=0.00000391$

The p value is very low compared to the significance level of $\alpha=0.05$ then we can reject the null hypothesis and we can conclude that the true proportion of people liberal is higher than 0.24

Step-by-step explanation:

Information given

n=200 represent the random sample taken

X=75 represent the number of people Liberal

$\hat p=\frac{75}{200}=0.375$ estimated proportion of people liberal

$p_o=0.24$ is the value that we want to test

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to verify if the true proportion of adults liberal is higher than 0.24:

Null hypothesis: $p \leq 0.24$

Alternative hypothesis: $p > 0.24$

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:

$z=\frac{0.375 -0.24}{\sqrt{\frac{0.24(1-0.24)}{200}}}=4.47$

Now we can calculate the p value based on the alternative hypothesis with this probability:

$p_v =P(z>4.47)=0.00000391$

The p value is very low compared to the significance level of $\alpha=0.05$ then we can reject the null hypothesis and we can conclude that the true proportion of people liberal is higher than 0.24

6 0

3 years ago