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Arlecino [84]
2 years ago
10

The mean of children in 5 families is 3. How many children might be in each family if none of them have 3 children?? ahh please

help this was due two days ago ill give brainliestttt​
Mathematics
1 answer:
Nadusha1986 [10]2 years ago
4 0

Answer:

draw a tree diagram to determine the sample space Step-by-step explanation:

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FIND THE GCF. Mathematics
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The G.C.F of the given algebraic expression is; ⁴⁹/₆a xy

<h3>What is the G.C.F (Greatest Common Factor)?</h3>

The greatest common factor (GCF or GCD or HCF) of a set of whole numbers is the largest positive integer that divides evenly into all numbers with zero remainder. For example, for the set of numbers 18, 30 and 42 the GCF = 6.

We are given the algebraic expression;

(3 x y * 4x²y * ¹/₁ * 5x³y ^ z * 4)7a * 1/1 * 7*6

Expanding this further gives;

⁴⁹/₆a((3xy * 4x²y * 20x³y ^ z )

Now, the GCF of the terms inside the bracket would be x y. Thus, expanding the GCF  we have;

⁴⁹/₆a xy((3 * 4x * 20x²y^(z - 1))

Read more about G.C.F at; brainly.com/question/219464

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1 year ago
WHOEVER ANSWERS THE FASTEST WINS BRAINIEST ANSWER AND GETS 98 POINTS FOR THIS QUESTION!!!!!! WHO CAN DO IT?
olga nikolaevna [1]

coordinates of b'a'c are

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B:(-5,10)

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3 0
2 years ago
Read 2 more answers
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.

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6 0
3 years ago
Which inequality is the solution to 20 + 2 &lt; x/3 - 8?
pychu [463]

Answer:

x > 90, C

Step-by-step explanation:

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