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LenaWriter [7]
2 years ago
11

Barney ate 21 M&M's. If he ate 15% of the bag of M&M's, how many were in the bag to start?

Mathematics
2 answers:
natali 33 [55]2 years ago
8 0

Answer:

3.15

i just used the calculater :)

ryzh [129]2 years ago
3 0

Answer:

There were 140 M&M's in the bag to start.

Step-by-step explanation:

Note: The word "of" is a good indicator of multiplication in word problems.

Let's say x is the number of M&M's in the bag. Then,

21=x*15%

Solve for x by dividing by 15% on both sides

140=x

There were 140 M&M's in the bag to start.

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Answer:

Yes it is possible

Step-by-step explanation:

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3 years ago
−7x &gt; 4(−3x − 5)<br><br> I'm bad at math
VashaNatasha [74]

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Hope that helps

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6 0
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VashaNatasha [74]
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4 0
3 years ago
A theory predicts that the mean age of stars within a particular type of star cluster is 3.3 billion years, with a standard devi
Luden [163]

Answer:

Null hypothesis:\mu \leq 3.3  

Alternative hypothesis:\mu > 3.3  

z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768  

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

p_v =P(z>1.768)=0.039  

Step-by-step explanation:

Data given and notation

\bar X=3.4 represent the sample mean  

\sigma=0.4 represent the population deviation for the sample

n=50 sample size  

\mu_o =3.3 represent the value that we want to test  

\alpha 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 to be tested  

We need to conduct a hypothesis in order to determine if the mean is higher than 3.3, the system of hypothesis would be:  

Null hypothesis:\mu \leq 3.3  

Alternative hypothesis:\mu > 3.3  

Compute the test statistic  

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-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".  

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

z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768  

P value

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

p_v =P(z>1.768)=0.039  

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Answer:

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