He lifted 100 kilograms and 132.5 kilograms. What was the combined weight of his lifts in grams?

A dolphin jumps 54 times in a 6 hour period. At this rate, how many jumps would be expected in a 2-hour period?

Lemur [1.5K]

Answer:

The dolphin will jump 18 times in 2hours.

Step-by-step explanation:

Find the jumps per hour: 54/6 = 9

^This means that the dolphin jumps 9 times in an hour.

9*2 (hours) = 18

The dolphin will jump 18 times in two hours.

3 0

2 years ago

Help with question 2

ANTONII [103]

The area can be found by multiplying the two together. 11x12=132.

The area is 132.

5 0

3 years ago

Read 2 more answers

8 friends share 1/3 of a bag of chips. If they each get an equal share how much of the bag of chips does each friend get?

SVETLANKA909090 [29]

Answer:

0.04166666666

I think, im not sure

Step-by-step explanation:

1/3 divided 8 = 0.417 (rounded)

5 0

2 years ago

The radius of a circle is 15. Using π, which equation expresses the ratio of the circumference to the diameter of a circle?

Kobotan [32]

Answer:

The equation which expresses the ratio of the circumference to the diameter of a circle is $\frac{C}{30}=\pi$ ⇒ D

Step-by-step explanation:

π is the ratio between the circumference of the circle and the length of its diameter

$\frac{C}{d}=\pi$ , where

C is the circumference of the circle
d is the diameter of the circle

∵ The circumference of the circle is C

∵ The radius of the circle is 15 units

∴ r = 15 units

- Find the diameter of the circle

∵ The diameter of the circle d = 2 r

∴ d = 2(15) = 30 units

∴ The diameter of the circle is 30 units

∵ $\frac{C}{d}=\pi$

∵ d = 30 units

∴ $\frac{C}{30}=\pi$

The equation which expresses the ratio of the circumference to the diameter of a circle is $\frac{C}{30}=\pi$

6 0

4 years ago

An e-commerce research company claims that 60% or more graduate students have bought merchandise on-line at their site. A consum

Elena-2011 [213]

Answer:

$z=\frac{0.55 -0.6}{\sqrt{\frac{0.6(1-0.6)}{80}}}=-0.913$

$p_v =P(z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of graduate students show that only 44 students have ever done so is not significantly lower than 0.6

Step-by-step explanation:

Data given and notation

n=80 represent the random sample taken

X=44 represent the students that have bought merchandise on-line at their site

$\hat p=\frac{44}{80}=0.55$ estimated proportion of graduate students show that only 44 students have ever done so

$p_o=0.6$ is the value that we want to test

$\alpha$ represent the significance level

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 interest is lower than 0.6 or 60%, so then the system of hypothesis are.:

Null hypothesis: $p \geq 0.6$

Alternative hypothesis: $p < 0.6$

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 $\hat p$ 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:

$z=\frac{0.55 -0.6}{\sqrt{\frac{0.6(1-0.6)}{80}}}=-0.913$

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 assumed for this case is $\alpha=0.05$ . 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 very high value and using the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of graduate students show that only 44 students have ever done so is not significantly lower than 0.6

5 0

3 years ago