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DedPeter [7]
2 years ago
6

Hey I need some help!! Does anyone mind giving me the answers and explain how? I really don’t understand. PS the question is in

the pic, I will mark u brainlist if you help!! Have a wonderful day everyone!!!!! Technically I’m supposed to match them btw!

Mathematics
1 answer:
Ksenya-84 [330]2 years ago
6 0

Answer:

2x2x2x5

Step-by-step explanation:

2x2x2x5 equals 40

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Find the greatest common factor of 12and 36
Xelga [282]
12:(3)(2)(2)
36:(3)(3)(2)(2)
GCF:12
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I need to know the answer to this problem plz
jekas [21]
Yes because 6×6 is 36 and 9×4 is 36.
the blue shows who will get the free apple juice and the red the free popcorn the last one is the 36th person

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a survey mr thrones class shows that 5 out of 8 students will by lunch how many out of 720 students in the school will buy lunch
never [62]

5/8 that's how ratio of students that will buy lunch. so in order to find number students who will buy the lunch out of 720, just multiply 720 with 5/8 ie, 720x5/8 ie 450.

450 is answer

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2 years ago
Assume that you have a sample of n 1 equals 6​, with the sample mean Upper X overbar 1 equals 50​, and a sample standard deviati
tigry1 [53]

Answer:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

df=6+5-2=9

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

n_1 =6 represent the sample size for group 1

n_2 =5 represent the sample size for group 2

\bar X_1 =50 represent the sample mean for the group 1

\bar X_2 =38 represent the sample mean for the group 2

s_1=7 represent the sample standard deviation for group 1

s_2=8 represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 > \mu_2

We are assuming that the population variances for each group are the same

\sigma^2_1 =\sigma^2_2 =\sigma^2

The statistic for this case is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

The pooled variance is:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

We can find the pooled variance:

S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67

And the pooled deviation is:

S_p=7.46

The statistic is given by:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

The degrees of freedom are given by:

df=6+5-2=9

The p value is given by:

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

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3 years ago
The equation with one change and no solutions: 8x + 6 =
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The answer is 2 ........
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