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3 years ago

Which is the best estimate of 174% of 141?

Mathematics

2 answers:

Vaselesa [24]3 years ago

7 0

Answer:

ITS.................................. 245

Step-by-step explanation:

xeze [42]3 years ago

4 0

245.34 is 174% of 141 :)
(1.74*141=245.34)

It helps to remember that "of" in math means to multiply. That's helped me for years.

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mark ordere wome balloons for a party that she is having.The balloons cost $4.50 each and there is a service fee of $12.The tota

ohaa [14]

A general equation to use for this situation is y = mx + b.

For this question, we can assume that y is total cost, m is cost per balloon, x is the amount of balloons, and b as the service fee; so we can set the equation up:

y = (4.50)x + 12

And we can further plug in the total cost to find the number of balloons purchased for the party:

79.50 = (4.50)x + 12

Now we can solve for x (number of balloons):

67.50 = (4.50)x

x = 15

The total number of balloons purchased for the party is 15.

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3 years ago

If the sphere shown has a radius of 9 units, what is the volume of the sphere?

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3 years ago

25m/s to miles/h Can anyone solve it? With explanation if you can

Dimas [21]

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round answer to either 55.9 or 56 mph depending on what you need.

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3 years ago

Write the below algebraic expression in words 3x+ (x)=10 2

enyata [817]

Answer:

Step-by-step explanation:

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3 years ago

A genetic experiment involving peas yielded one sample of offspring consisting of 420 green peas and 174 yellow peas. Use a 0.01

slavikrds [6]

Answer:

a) $z=\frac{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649$

b) For this case we need to find a critical value that accumulates $\alpha/2$ of the area on each tail, we know that $\alpha=0.01$ , so then $\alpha/2 =0.005$ , using the normal standard table or excel we see that:

$z_{crit}= \pm 2.58$

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

Step-by-step explanation:

Data given and notation

n=420+174=594 represent the random sample taken

X=174 represent the number of yellow peas

$\hat p=\frac{174}{594}=0.293$ estimated proportion of yellow peas

$p_o=0.23$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

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 of yellow peas is 0.23:

Null hypothesis: $p=0.23$

Alternative hypothesis: $p \neq 0.23$

When we conduct a proportion test we need to use the z statisitc, 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 $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

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

$z=\frac{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649$

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 bilateral test the p value would be:

$p_v =2*P(z>3.649)=0.00026$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

b) Critical value

For this case we need to find a critical value that accumulates $\alpha/2$ of the area on each tail, we know that $\alpha=0.01$ , so then $\alpha/2 =0.005$ , using the normal standard table or excel we see that:

$z_{crit}= \pm 2.58$

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

5 0

3 years ago