Which expression is equivalent to 56 + 21?

You could win a giant jar of M&Ms, if only you could guess how many are in the jar. You find a smallerjar such that the gian

sleet_krkn [62]

Answer:

There are 7360 M&Ms in larger jar.

Step-by-step explanation:

We are given the following in the question:

A giant jar is such that it is exactly four times times the smaller jar in dimensions.

If r and h is the radius and height of the cylindrical jar respectively, then the dimensions of the larger are:

$R = 4r\\H=4h$

Number of M&Ms in smaller jar = 115

Volume of smaller jar =

$v = \dfrac{1}{3}\pi r^2 h$

Volume of larger jar =

$V =\dfrac{1}{3}\pi R^2 H=\dfrac{64}{3}\pi r^2 h$

Number of M&Ms in larger jar:

$v = \dfrac{1}{3}\pi r^2 h\rightarrow 115\\\\V=\dfrac{64}{3}\pi r^2 h\rightarrow 115\times 64\\\\V=\dfrac{64}{3}\pi r^2 h\rightarrow 7360$

Thus, there are 7360 M&Ms in larger jar.

6 0

3 years ago

What is meant by core fitness

solniwko [45]

Core fitness refers to exercises to activate the muscles in a location called the core of your body. Hope this helps!

5 0

3 years ago

HELP !! Super confused

OleMash [197]

Answer:

ik the answer to the question

3 0

3 years ago

A researcher believes that the mean weight of competitive runners is about 140 pounds. A sample of 24 elite distance runners has

ExtremeBDS [4]

Answer:

$t=\frac{136-140}{\frac{11}{\sqrt{24}}}=-1.781$

$p_v =P(t_{(23)}$

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is less than 140 pounds at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=136$ represent the sample mean

$s=11$ represent the sample standard deviation

$n=24$ sample size

$\mu_o =140$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 140, the system of hypothesis would be:

Null hypothesis: $\mu \geq 140$

Alternative hypothesis: $\mu < 140$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{136-140}{\frac{11}{\sqrt{24}}}=-1.781$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=24-1=23$

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

$p_v =P(t_{(23)}$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is less than 140 pounds at 5% of signficance.

4 0

3 years ago

Three boxes contain counters

Strike441 [17]

What we Know: 3 box have some counters. There are 63 counters when they are all combined.
A and B counters = 36
The difference between A and C is 10.
Since we already know A and B we can subtract them from 63 to get C
C= 63- 36 = 27
Now we have to find A by looking for what number gives us the difference of 10 when subtracted from C.
36 = 18+ 18 or 19+ 17
18 and 18 wont work because it does not give us the difference of 10.
So A = 17 counters and B = 19 counters.

3 0

3 years ago