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BaLLatris [955]
3 years ago
5

Miriam bought a bushel of apples from the apple orchard to bake apple pies for a bake sale. She started with 126 apples. After t

he first day of baking, she had 108 apples left. After the second day, she had 90 apples. After the third day, she had 72 apples left.If she continues this pattern, how many apples will she have left after 7 days of baking?
Mathematics
2 answers:
natulia [17]3 years ago
8 0
Miriam will have no apples left after the 7th day of baking.

Here is the number of apples she will have after each day starting with after the 1st day of baking.

126, 108(1st day), 90(2nd day), 72(3rd day), 54(4th day), 36(5th day), 18(6th day), 0(7th day).

Each day the number of apples decreases by 18.
Marianna [84]3 years ago
3 0

Arithmetic, because the sequence has a common difference. answer on edgenui......

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students who failed mathematics scored the following marks 16.5,13.6,12.4,13.7 find the average mark for this students​
SSSSS [86.1K]

Answer:

16.5+13.6+12.4.17.7= 60.2

60.2÷4 =15.05

8 0
3 years ago
1)cot a/2 -tan a/2 = 2 cot a<br>2) cot b/2 + tan b/2= 2 cosec b<br> prove​
VladimirAG [237]
1)

LHS = cot(a/2) - tan(a/2)

= (1 - tan^2(a/2))/tan(a/2)

= (2-sec^2(a/2))/tan(a/2)

= 2cot(a/2) - cosec(a/2)sec(a/2)

= 2(1+cos(a))/sin(a) - 1/(cos(a/2)sin(a/2))

= 2 (1+cos(a))/sin(a) - 2/sin(a)) (product to sums)

= 2[(1+cos(a) -1)/sin(a)]

=2cot a

= RHS

2.

LHS = cot(b/2) + tan(b/2)

= [1 + tan^2(b/2)]/tan(b/2)

= sec^2(b/2)/tan(b/2)

= 1/sin(b/2)cos(b/2)

using product to sums

= 2/sin(b)

= 2cosec(b)

= RHS
8 0
2 years ago
4-(3-5)=<br> I need help
liraira [26]

Answer:

6

Step-by-step explanation:

I'm 99% sure you know the answer to this and didn't have to resort to this website.

3-5 = -2

Subtracting negative numbers means it becomes positive.

So the gist is that your adding 4 and 2 which is 6.

6 0
3 years ago
A random sample of 25 ACME employees showed the average number of vacation days taken during the year is 18.3 days with a standa
Norma-Jean [14]

Answer:

a) Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

b) df=n-1=25-1=24  

For this case the p value is given p_v = 0.0392

If we compare the p value and the significance level given \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 true mean is higher than 15 at 5% of signficance.  

c) Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

Step-by-step explanation:

Data given and notation  

\bar X=18.3 represent the sample mean

s=3.72 represent the sample standard deviation

n=25 sample size  

\mu_o =15 represent the value that we want to test

\alpha=0.05 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)  

Part a: State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean for vacation days is higher than 15, the system of hypothesis would be:  

Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

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".  

Part b: P-value  and conclusion

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

df=n-1=25-1=24  

For this case the p value is given p_v = 0.0392

Conclusion  

If we compare the p value and the significance level given \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 true mean is higher than 15 at 5% of signficance.  

Part c

Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

3 0
3 years ago
Can you please help???
marusya05 [52]

Answer:

2 or 3 I think

I gessssssssssing

5 0
2 years ago
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