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3 years ago

8+(-6x42)/497+42 Find the GCF

Mathematics

1 answer:

adoni [48]3 years ago

4 0

Answer:

49 35/71

Step-by-step explanation:

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You are conducting a study to test whether there is a difference in the mean number of alcoholic drinks consumed per evening bet

salantis [7]

The value of test statistic is 5.45

Data;

n1 = 25
n2 = 30
x1 = 6.2
x2 = 3.8
s1 = 2
s2 = 1

<h3>Test Statistic</h3>

The formula to calculate the test statistic is given as

$t = \frac{x_1 - x_2}{\sqrt{\frac{s_1^2}{n_1}+ \frac{s_2^2}{n_2} } } \\$

Let's substitute the values and solve for t

$t = \frac{x_1 - x_2}{\sqrt{\frac{s_1^2}{n_1}+ \frac{s_2^2}{n_2} } } \\t = \frac{6.2 - 3.8}{\sqrt{\frac{2^2}{25} + \frac{1^2}{30} } } \\t = \frac{2.4}{0.44} = 5.45$

The value of test statistic is 5.45

Learn more on test statistic here;

brainly.com/question/15980493

5 0

3 years ago

Help me with this question, please!! I don't understand!!

Gwar [14]

Answer:

Presentation on theme: "Geometry Spheres CONFIDENTIAL."— Presentation transcript:

Step-by-step explanation:

Search this up and watch the video it may explain it to you better. I hope this helps!

6 0

3 years ago

Can you please help me do this?

Evgen [1.6K]

Answer: y = 7

Have a great day!

7 0

3 years ago

Read 2 more answers

Assume that when adults with smartphones are randomly selected, 46% use them in meetings or classes. If 9 adult smartphone use

valina [46]

Answer:

0.18173219.

Step-by-step explanation:

We have been asked to find what will be the probability that at least 6 of 9 adults use their smartphones in meetings or classes.

We will find our answer using Bernoulli's trails.

$_{r}^{n}\textrm{c}\cdot p^{r}\cdot (1-p)^{n-r}$

First of all we will find the probabilities when r is 6, 7,8 and 9 then we will add them all.

When r=6,

$_{6}^{9}\textrm{c}\cdot 0.46^{6}\cdot (1-0.46)^{9-6}$

$\frac{9!}{6!3!} *0.46^{6} *0.54^{3}$

$\frac{9*8*7*6!}{6!*3*2*1} *0.009474296896*0.157464$

$84*0.009474296896*0.157464=0.12532$

Similarly we will find Probabilities when r=7, 8 and 9.

When r=7

$_{7}^{9}\textrm{c}\cdot 0.46^{7}\cdot (1-0.46)^{9-7}$

$\frac{9!}{7!2!} *0.46^{7} *0.54^{2}$

$\frac{9*8*7!}{7!*2*1} *0.00435817657216*0.2916$

$36*0.00435817657216*0.2916=0.04575039$

When r=8

$_{8}^{9}\textrm{c}\cdot 0.46^{8}\cdot (1-0.46)^{9-8}$

$\frac{9!}{8!1!} *0.46^{8} *0.54$

$\frac{9*8!}{8!*1!} *0.00200476*0.54$

$9 *0.00200476*0.54=0.0097431336$

When r=9,

$_{9}^{9}\textrm{c}\cdot 0.46^{9}\cdot (1-0.46)^{9-9}$

$\frac{9!}{9!0!} *0.46^{9} *1$

$1 *0.00092219*1=0.00092219$

Now let us add all the probabilities to get the final answer.

$0.12532+0.04575+0.00974+0.00092219=0.18173219$

Therefore, probability that at least 6 of 9 adults use their smartphones in meetings or classes is 0.18173219.

6 0

4 years ago

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Work the following pricing problems for services rendered. (For all calculations use hundredths.)

Bogdan [553]

Answer:

No entiendo ingles perdon

4 0

3 years ago

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