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Ivan
3 years ago
9

The product of two consecutive odd integers is 255. Find the integers.

Mathematics
1 answer:
Grace [21]3 years ago
5 0

Answer:

255 5,6 7,78

Step-by-step explanation:

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Need help on this question asap please
Natalka [10]

Answer:

a= 168

p= 62

Step-by-step explanation:

25 24 and 7 are pythagorean triples. meaning the short side length is 7.

24*7 = 168

48+14=62

5 0
3 years ago
Describe the solutions of 5
djyliett [7]

Answer:

A value, or values, we can put in place of a variable (such as x) that makes the equation true.

Example: x + 2 = 7

When we put 5 in place of x we get: 5 + 2 = 7

5 + 2 = 7 is true, so x = 5 is a solution

6 0
3 years ago
Which is equivalent to 5(a + 6)?<br><br> 1. a+30<br><br> 2. 5a-30<br><br> 3. 5a+6<br><br> 4. 5a+30
Afina-wow [57]

Answer:  the answer is number four.

Step-by-step explanation:

<h2>because when you take five multiply by A (5*a) you will get 5a and when you take five multiply by six (5*6) you will get 30 and there is positive sign so the plus (+) stays the same and your will be (5a+30) that is your answer. good luck.</h2>
5 0
3 years ago
What is a typical age (the mean) for the people who were at Movie A?
Snowcat [4.5K]

It is 39

Because you have to count the little dots all the way till it is around the most age.

7 0
2 years ago
Nationwide, it is estimated that 40% of service stations have gas tanks that leak to some extent. A new program in California is
fredd [130]

Answer:

Since the calculated value of z= -1.496 does not fall in the critical region z < -1.645 we conclude that the new program is effective. We fail to reject the null hypothesis .

Step-by-step explanation:

The sample proportion is p2= 7/27= 0.259

and q2= 0.74

The sample size = n= 27

The population proportion = p1= 0.4

q1= 0.6

We formulate the null and alternate hypotheses that the new program is effective

H0: p2> p1   vs  Ha: p2 ≤ p1

The test statistic is

z= p2- p1/√ p1q1/n

z= 0.259-0.4/ √0.4*0.6/27

z=  -0.141/0.09428

z= -1.496

The significance level ∝ is 0.05

The critical region for one tailed test is z ≤ ± 1.645

Since the calculated value of z= -1.496 does not fall in the critical region z < -1.645 we conclude that the new program is effective. We fail to reject the null hypothesis .

8 0
3 years ago
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