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Grace [21]
4 years ago
6

What are three integers that do not all have the same sign that have a sum of -20. Write three integers that do not all have the

same sign that have a sum of 10
Mathematics
1 answer:
Ratling [72]4 years ago
8 0

There isn't an unique solution to these equations. For example, for the first one, you may pick two arbitrary positive number, and fix the third so that the sum is -20.

Say that you choose 4 and 26. Their sum is 30. To reach -20, the third number must be -50.

So, 4, 26 and -50 are three numbers that sum to -20 and don't have all the same sign.

Similarly, for the second requirement, pick two negative numbers, and adjust the third.

If you choose -3 and -2, their sum is -5. So, the third must be 15. In fact, -3, -2 and 15 are three numbers that sum to 10 and don't have all the same sign.

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Four more than the quotient of a number and 3 is 2
Arturiano [62]

Answer:4+x/3=2

Step-by-step explanation:

4 more=4+

the quotient of a number and 3= x/3

the anwser = =2

6 0
3 years ago
Brent is a researcher for a food company. He is on a team creating a reduced-calorie version of its flagship cracker. The team w
Andre45 [30]

Answer:

Step-by-step explanation:

Hello!

The research team created a cracker with fewer calories. The average content of calories of the new crackers per serving of 6 should be less than 60.

To test it a random sample of 26 samples of the new cracker was taken and the calories per serving were measured.

Then the study variable is

X: calories of a 6 serve sample of the new reduced-calorie version. (cal)

The variable has a normal distribution with a population standard deviation of 0.82 cal.

To test the claim that the new crackers have on average less than 60 calories, the parameter of interest is the population mean (μ) and the hypotheses are:

H₀: μ ≥ 60

H₁: μ < 60

α: 0.01

Since the variable has a normal distribution and the population variance is known, the best statistic to use to conduct the test is a Standard Normal

Z= \frac{(X[bar]-Mu)}{\frac{Sigma}{\sqrt{n}}  } ~N(0;1)

This test is one tailed to the left, wich means that the null hypothesis will be rejected at low levels of the statistic.

Z_{\alpha } = Z_{0.01} = -2.334

If Z ≤ -2.334, the decision is to reject the null hypothesis.

If Z > -2.334, the decision is to not reject the null hypothesis.

Using the data of the sample I've calculated the sample mean.

X[bar]= ∑X/n= 1548.61/26= 59.56 cal

Z_{H_0}= \frac{(59.56-60)}{\frac{0.82}{\sqrt{26} } } = -2.736

The observed Z value is less than the critical value, so the decision is to reject the null hypothesis.

At a level of significance of 1%, you can conclude that the population mean of calories of the samples of new crackers is less than 60 cal.

I hope it helps!

6 0
3 years ago
If one root of the equation is 4x^2- 2x+p-4 be the reciprocal of other. then find the value of p​
SashulF [63]

Answer:

p = 8

Step-by-step explanation:

Let one root of the eqn. be alpha . Other root is 1/alpha .

We know that product of both roots of an quadratic eqn. is c/a where "c" is the co-efficient of the constant  & "a" is the co-efficient of x^2.

Here "c" is p-4 & "a" is 4. And the product of roots is 1 ( ∵ prdouct of a number and its reciprocal is 1 )

\frac{c}{a} = 1\\=> \frac{p - 4}{4} = 1\\=> p - 4 = 4\\=> p = 4 + 4 = 8

5 0
4 years ago
How should I study for Regents Living Environment Exam?
shtirl [24]
You should try reading notes you took during your class period.
5 0
3 years ago
A boat travels 12 km upstream and back in 1 hour 45 minutes. If the speed of the current is 3 km/h throughout, find the speed of
Elza [17]

Answer:

14.3 km/h   to 3 significant figures

Step-by-step explanation:

Let time travelling upstream be t1 hours and downstream = t2 hours.

t1 + t2 = 1.75..........................(1)

Let v  be the speed of the boat in still water.

12 / t1 = v - 3.........................(2)

12 /  t2 = v + 3........................(3)

From (1) and (3) , substituting for t2:-

12 / (1.75 - t1) = v + 3 ..................(4)

Subtract:-   (2) - (4):-

12 / t1 - 12 / (1.75 - t1) = -6

12(1 .75 - t1) - 12t1 = -6t1(1.75 -  t1)

21 - 12t1 - 12t1 = -10.5t1 + 6t^2

6t1^2 + 13.5t1 - 21 = 0

Solving this gives t1 = 1.058 hours

Therefore the speed in still water (v)  = 12/1.058 + 3 = 14.3 to 3 sig figs

3 0
3 years ago
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