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nydimaria [60]
4 years ago
12

Write a pair of integers whose difference gives. (a) a negative integer. (b) zero. (c) an integer smaller than both the integer.

(d) an integer smaller than only one of the integers. (e) an integer greater than both the integers.
Mathematics
1 answer:
chubhunter [2.5K]4 years ago
5 0

Answer:

(a)5 and 8

(b)5 and 5

(c) 7 and 5

(d)6 and 2

(e)6 and -2

Step-by-step explanation:

(a)a negative integer

We consider the integers 5 and 8

Difference = 5-8 =-3

3 is a negative integer.

(b)Zero

We consider the integers 5 and 5

Difference = 5-5 =0

(c) an integer smaller than both the integer.

We consider the integers 7 and 5

Difference = 7-5 =2

2 is smaller than both integers.

(d) an integer smaller than only one of the integers.

We consider the integers 6 and 2

Difference =6-2=4

4 is smaller than only 6.

(e)an integer greater than both the integers

We consider the integers 6 and -2

Difference =6-(-2)=6+2=8

8 is greater than both 6 and -2.

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There are ten people in the Baking Club, including Mark. They choose $3$ people to form an executive committee. How many possibl
r-ruslan [8.4K]

Answer:

120 ways

Step-by-step explanation:

We have the total number of people in this baking club = 10, Mike inclusive.

We are asked to find how to form an executive committee of 3 from these 10 people

We solve this using combination method

10C3

= 120 possible committees

So out of these 10 people we can select a committee of 3 persons in 120 ways.

Thank you.

6 0
3 years ago
Question 1 options:Residents in Portland, Oregon think that their city has more rainfall than Seattle, Washington. To test this
dimaraw [331]

Answer:

There is no enough evidence to to support the claim that Portland has more average yearly rainfall than Seattle.

Being μ1: average rainfall in Portland, μ2: average rainfall in Seattle, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 > 0

P-value = 0.1290

As the P-value is bigger than the significance level, the effect is not significant and the null hypothesis failed to be rejected.

Step-by-step explanation:

We have to test the hypothesis of the difference between means.

The claim is that Portland has more average yearly rainfall than Seattle.

Being μ1: average rainfall in Portland, μ2: average rainfall in Seattle, the null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 > 0

The significance level is 0.10.

The sample for Portland, of size n1=45, has a mean of M1=37.50 and standard deviation of s1=1.82.

The sample for Seattle, of size n1=35, has a mean of M1=37.07 and standard deviation of s1=1.68.

The difference between means is:

M_d= M_1-M_2=37.50-37.07=0.43

The standard error for the difference between means is:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{1.82^2}{45}+\dfrac{1.68^2}{35}}=\sqrt{ 0.0736+0.0688 }=\sqrt{0.1424}\\\\\\s_{M_d}=0.3774

We can calculate the t-statistic as:

t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.43-0}{0.3774}=1.1393

The degrees of freedom are:

df=n1+n2-2=45+35-2=78

Then, the p-value for this one-tailed test with 78 degrees of freedom is:

P-value=P(t>1.1393)=0.1290

As the P-value is bigger than the significance level, the effect is not significant and the null hypothesis failed to be rejected.

There is no enough evidence to to support the claim that Portland has more average yearly rainfall than Seattle.

7 0
3 years ago
Kyle spends $44 for four sweatshirts.
Arturiano [62]

Answer:

I think the answer is B. sorry if its wrong..

Step-by-step explanation:

6 0
3 years ago
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SOLVE FOR 50 POINTS CORRECTLY! OR ELSE REPORT!
Trava [24]
Y=16.65
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4 0
2 years ago
Please help!
ollegr [7]
The possible value of b is:
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6 0
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