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zheka24 [161]
3 years ago
11

For a class experiment, Sally's class weighed a log before and after

Mathematics
1 answer:
timama [110]3 years ago
3 0

Answer:

Jack Barlow

Step-by-step explanation:

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What is the value of the fourth term in a geometric sequence for which a=15 and r= 1/3
Andreyy89

Answer:

Step-by-step explanation:

hello :

the n-ieme term is : An=A1×r^(n-1)

A1 the first term       r : the common ratio

in this exercice : A1 =15        r = 1/3      n = 4

A4=15×(1/3)^(4) =15×4^4 =3840    

6 0
2 years ago
What is the equation of the line that has a slope of
sveticcg [70]
The answer is D.

If the slope is -4/5 and the y int. is -1/6 then x is the slope.
4 0
2 years ago
A survey of 90 men found that an average amount spent on St. Patrick's day of $55 with a standard deviation of $18. A similar su
dem82 [27]

Answer:

The value of the test statistic is 4.70.

Step-by-step explanation:

The hypothesis for this test can be defined as follows:

<em>H</em>₀: Men do not spend more than women on St. Patrick's day, i.e. μ₁ = μ₂.

<em>H</em>ₐ: Men spend more than women on St. Patrick's day, i.e. μ₁ > μ₂.

The population standard deviations are not known.

So a <em>t</em>-distribution will be used to perform the test.

The test statistic for the test of difference between mean is:

t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s^{2}_{1}}{n_{1}}+\frac{s^{2}_{2}}{n_{2}}} }

Given:

\bar x_{1}=55\\s_{1}=18\\n_{1}=90\\\bar x_{1}=44\\s_{1}=16\\n_{1}=86

Compute the value of the test statistic as follows:

t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s^{2}_{1}}{n_{1}}+\frac{s^{2}_{2}}{n_{2}}} }=\frac{55-44}{\sqrt{\frac{15^{2}}{90}+\frac{16^{2}}{86} }}=4.70

Thus, the value of the test statistic is 4.70.

7 0
3 years ago
Alex has a pet snake that eats 1 mouse every week. How many mice will the snake eat if Alex has the snake for 4 years?
Anvisha [2.4K]

Answer:

208.5714

Step-by-step explanation:

how many weeks are in 4 years

3 0
3 years ago
Read 2 more answers
Several years​ ago, 38​% of parents who had children in grades​ K-12 were satisfied with the quality of education the students r
Colt1911 [192]

Answer:

0.428 - 1.96\sqrt{\frac{0.428(1-0.428)}{1165}}=0.3995

0.428 + 1.96\sqrt{\frac{0.428(1-0.428)}{1165}}=0.4564

We are confident that the true proportion of people satisfied with the quality of education the students receive is between (0.3995, 0.4564), since the lower value for this confidence level is higher than 0.38 we have enough evidence to conclude that the parents' attitudes toward the quality of education have changed.

Step-by-step explanation:

For this case we are interesting in the parameter of the true proportion of people satisfied with the quality of education the students receive

The confidence level is given 95%, the significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical values are:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The estimated proportion of people satisfied with the quality of education the students receive is given by:

\hat p =\frac{499}{1165}= 0.428

The confidence interval for the proportion if interest is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

And replacing the info given we got:

0.428 - 1.96\sqrt{\frac{0.428(1-0.428)}{1165}}=0.3995

0.428 + 1.96\sqrt{\frac{0.428(1-0.428)}{1165}}=0.4564

We are confident that the true proportion of people satisfied with the quality of education the students receive is between (0.3995, 0.4564), since the lower value for this confidence level is higher than 0.38 we have enough evidence to conclude that the parents' attitudes toward the quality of education have changed.

8 0
3 years ago
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