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pickupchik [31]
3 years ago
6

1. In a game, if you roll a 6 on a 6-sided number cube, you lose a turn.

Mathematics
2 answers:
USPshnik [31]3 years ago
8 0

Answer:

its 1/2

Step-by-step explanation:

i just took the test and the first wrong was wrong its not 20

(EDUNUITY)

just olya [345]3 years ago
3 0
Short answer = 20
Argument
Your chances of throwing a 6 = 1/6

Expected Throws = P(6) * Total Throws
Expected Throws = 1/6 * 120
Expected Throws = 20

Answer = 20 <<<<<
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The total age of a woman and her son is 51 years. Three years ago, the woman was eight times as old as her son. How is old is he
Veronika [31]

51 - 3 = 48

48 / 8 = 6

6 + 3 = 9

Her son is 9 years old

6 0
3 years ago
Write the following as an expression or equation <br> 10 less than the product of a number and four
ehidna [41]
10 - 4x

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A football player starts on the 20 yard line and gains 9 yards. On the next play, he loses 4 yards. The referee says there’s a p
julsineya [31]
The answer would be A because if the player starts on the 20 yard line and gains 9 yards, that would be 20+9= 29. Then it says the player loses 4 yards which would make it 29-4= 25 then it says the referee takes the penalty away which then makes it 29 again. If you do the math for A it comes out to be 29.
7 0
2 years ago
A line passes through the points (4, 9) and (12, 20). What is the slope of the line?
Mashutka [201]

Answer:

\frac{11}{8}

Step-by-step explanation:

→ State the gradient formula

( y₂ - y₁ ) ÷ ( x₂ - x₁ )

→ Find x₁ , x₂ , y₁ and y₂

x₁ = 4 , x₂ = 12 , y₁ = 9 and y₂ = 20

→ Substitute in the values

\frac{20-9}{12-4}

→ Simplify

\frac{11}{8}

4 0
2 years ago
A nutritionist has developed a diet that she claims will help people lose weight. Twelve people were randomly selected to try th
deff fn [24]

The diet which is developed by the nutritionist is effective at the 0.05 level of significance. Claim is agreed.

<h3>When do we use two-sample t-test?</h3>

The two-sample t-test is used to determine if two population means are equal.

A nutritionist has developed a diet that she claims will help people lose weight. In this,

  • Twelve people were randomly selected to try the diet.
  • Their weights were recorded prior to beginning the diet and again after 6 months.

Here are the original weights, in pounds, with the weight after 6 months in parentheses.

  • Before 192 212 171 215 180 207 165 168 190 184 200 196
  • After    183 196  174 211 160 191   162 175 190  179  189 195

The mean of the weights before 6 moths is,

\overline X_1=\dfrac{192+ 212 +171 +215 +180 +207 +165 +168 +190 +184 +200 +196 }{12}\\\overline X_1=190

The mean of the weights after 6 months is,

\overline X_2=\dfrac{ 183 +196  +174 +211 +160 +191   +162 +175 +190  +179  +189 +195  }{12}\\\overline X_1=183.75

Standard deviation of both the data is 16.9 and 14.7.

1. Null and Alternative Hypotheses.

The following null and alternative hypotheses need to be tested:

\begin{array}{ccl} H_0: \mu_1 & = & \mu_2 \\\\ \\\\ H_a: \mu_1 & > & \mu_2 \end{array}

This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

  • (2) Rejection Region

Based on the information provided, the significance level is α=0.05 and the degrees of freedom are df = 22. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

df_{Total} = df_1 + df_2 = 11 + 11 = 22

Hence, it is found that the critical value for this right-tailed test is

t_c=1.717, for α=0.05 and df=22

The rejection region for this right-tailed test is,

R = \{t: t > 1.717\}

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

t = \displaystyle \frac{\bar X_1 - \bar X_2}{\sqrt{ \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}(\frac{1}{n_1}+\frac{1}{n_2}) } }

\displaystyle \frac{ 190 - 183.75}{\sqrt{ \frac{(12-1)16.9^2 + (12-1)14.7^2}{ 12+12-2}(\frac{1}{ 12}+\frac{1}{ 12}) } } = 0.967

  • (4) Decision about the null hypothesis

Since it is observed that t=0.967≤tc=1.717 it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.1721 and since p=0.1721≥0.05p = 0.1721  it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis H₀ is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is greater than μ2​, at the α=0.05 significance level.

Confidence Interval

The 95% confidence interval is −7.16<μ<19.66

Thus, the diet which is developed by the nutritionist is effective at the 0.05 level of significance. Claim is agreed.

Learn more about the two-sample t-test here;

brainly.com/question/27198724

#SPJ1

8 0
2 years ago
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