Ask question

Ask question

All categories

3 years ago

15

The graph shows how the number of keychains in your collection increased since you began collecting five months ago. How many ke

ychains did you collect during the third month? PLZ HELP

2 answers:

lina2011 [118]3 years ago

7 0

Ten keychains in 3 months

prisoha [69]3 years ago

5 0

Answer:

fourth month

Step-by-step explanation:

You might be interested in

12 times 12 help please so work

Molodets [167]

144, its just repeated addition... 12+12+12+12+12+12+12+12+12+12+12+12=144

8 0

2 years ago

Read 2 more answers

Eight avocados cost $4. How much do 9 avocados cost?<br><br> i am dumb

blagie [28]

I believe the answer is 4.50

5 0

3 years ago

(20 points) A statistics practitioner calculated the mean and the standard deviation from a sample of 400. They are x = 98 and s

Nimfa-mama [501]

Answer:

The P-value is 0.0234.

Step-by-step explanation:

We are given that a statistics practitioner calculated the mean and the standard deviation from a sample of 400. They are x = 98 and s = 20.

Let = population mean.

So, Null Hypothesis, : = 100 {means that the population mean is equal to 100}

Alternate Hypothesis, : > 100 {means that the population mean is more than 100}

The test statistics that will be used here is One-sample t-test statistics because we're yet to know about the population standard deviation;

T.S. = ~

where, = sample mean = 98

s = sample standard deviation = 20

n = sample size = 400

So, the test statistics = ~

= -2

The value of t-test statistics is -2.

Now, the P-value of the test statistics is given by;

P( < -2) = 0.0234 {using the t-table}

8 0

3 years ago

Test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students at the 0.005 signif

algol [13]

Answer:

a) F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

b) B. two-tailed

d) $t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$

e) $p_v =P(t_{118}$

f) B. Reject the null hypothesis

Step-by-step explanation:

Information provided

$\bar X_{O}=2.91$ represent the mean for the Orange Coast

$\bar X_{C}=2.96$ represent the mean for the Coastline

$s_{O}=0.05$ represent the sample standard deviation for Orange Coast

$s_{C}=0.03$ represent the sample standard deviation for Coastline

$n_{O}=60$ sample size for Orange Coast

$n_{C}=60$ sample size for Coastline

$\alpha=0.005$ Significance level provided

t would represent the statistic

Part a

For this case we want to test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students

Null hypothesis: $\mu_{O} \geq \mu_{C}$

Alternative hypothesis: $\mu_{O} < \mu_{C}$

F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

Part b

For this case we need to conduct a left tailed test.

B. two-tailed

Part d

The statistic is given by:

$t=\frac{(\bar X_{O}-\bar X_{C})-\Delta}{\sqrt{\frac{s^2_{O}}{n_{O}}+\frac{s^2_{C}}{n_{C}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=60+60-2=118$

Replacing the info we got:

$t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$

Part e

We can calculate the p value with this probability:

$p_v =P(t_{118}$

Part f

Since the p value is a very low value compared to the significance level given of 0.005 we can reject the null hypothesis.

B. Reject the null hypothesis

5 0

2 years ago

A = 16 in , b = 8 in., h =

klasskru [66]

The height is 2 inches

7 0

3 years ago