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2 years ago

100 POINTS IF YOU ANSWER THIS QUESTION

Mathematics

2 answers:

Romashka [77]2 years ago

6 0

jones bbq foot massage

grandymaker [24]2 years ago

3 0

Answer:

if anyone is looking at this it's warm front is when a warm air mass slides over a cold air mass and stationary front is when a cold air mass and a warm air mass me and do not move and a cold front when a cold air mass slides under a warm air mass and it included front is when two cold fronts force a warm front up and block it from the surface

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Three coins are flipped. What is P(heads, heads, heads)?

pshichka [43]

1/8

Each time the probability for landing on heads in 1/2 or 50% so do 1/2*1/2*1/2=1/8.

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2 years ago

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61,229 rounding this

Lera25 [3.4K]

Are you rounding to the nearest 100’s place?

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2 years ago

A rhombus can belong to multiple categories. which catergories does this rhombus not belong to? Select the two correct answers.

amid [387]

Answer:

Im pretty sure its c and d

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2 years ago

To make a jug of plant fertilizer, Malaika uses 6 cups of water and 3 scoops of fertilizer. Bart uses 8 cups of water and 5 scoo

nexus9112 [7]

Answer:

No they will no have the same strength. So what I did was find a common number between them so between 8 and 6 they both can be multiplied to 24. So 6 goes into 24 4 times so I multiplied 3 times 6 and then 8 goes into 24 3 times so I multiplied 3 and 5 and so the final answer is that bart has more strong fertilizer

Step-by-step explanation:

6*4=24 4*3=12

8*3=24 5*3=15

5 0

2 years ago

A bank with a branch located in a commercial district of a city has the business objective of developing an improved process for

tatiyna

Answer:

(a) The test statistic value is -4.123.

(b) The critical values of t are ± 2.052.

Step-by-step explanation:

In this case we need to determine whether there is evidence of a difference in the mean waiting time between the two branches.

The hypothesis can be defined as follows:

H₀: There is no difference in the mean waiting time between the two branches, i.e. μ₁ - μ₂ = 0.

Hₐ: There is a difference in the mean waiting time between the two branches, i.e. μ₁ - μ₂ ≠ 0.

The data collected for 15 randomly selected customers, from bank 1 is:

S = {4.21, 5.55, 3.02, 5.13, 4.77, 2.34, 3.54, 3.20, 4.50, 6.10, 0.38, 5.12, 6.46, 6.19, 3.79}

Compute the sample mean and sample standard deviation for Bank 1 as follows:

$\bar x_{1}=\frac{1}{n_{1}}\sum X_{1}=\frac{1}{15}[4.21+5.55+...+3.79]=4.29$

$s_{1}=\sqrt{\frac{1}{n_{1}-1}\sum (X_{1}-\bar x_{1})^{2}}\\=\sqrt{\frac{1}{15-1}[(4.21-4.29)^{2}+(5.55-4.29)^{2}+...+(3.79-4.29)^{2}]}\\=1.64$

The data collected for 15 randomly selected customers, from bank 2 is:

S = {9.66 , 5.90 , 8.02 , 5.79 , 8.73 , 3.82 , 8.01 , 8.35 , 10.49 , 6.68 , 5.64 , 4.08 , 6.17 , 9.91 , 5.47}

Compute the sample mean and sample standard deviation for Bank 2 as follows:

$\bar x_{2}=\frac{1}{n_{2}}\sum X_{2}=\frac{1}{15}[9.66+5.90+...+5.47]=7.11$

$s_{2}=\sqrt{\frac{1}{n_{2}-1}\sum (X_{2}-\bar x_{2})^{2}}\\=\sqrt{\frac{1}{15-1}[(9.66-7.11)^{2}+(5.90-7.11)^{2}+...+(5.47-7.11)^{2}]}\\=2.08$

(a)

It is provided that the population variances are not equal. And since the value of population variances are not provided we will use a t-test for two means.

Compute the test statistic value as follows:

$t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}$

$=\frac{4.29-7.11}{\sqrt{\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}}}$

$=-4.123$

Thus, the test statistic value is -4.123.

(b)

The degrees of freedom of the test is:

$m=\frac{[\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}]^{2}}{\frac{(\frac{s_{1}^{2}}{n_{1}})^{2}}{n_{1}-1}+\frac{(\frac{s_{2}^{2}}{n_{2}})^{2}}{n_{2}-1}}$

$=\frac{[\frac{1.64^{2}}{15}+\frac{2.08^{2}}{15}]^{2}}{\frac{(\frac{1.64^{2}}{15})^{2}}{15-1}+\frac{(\frac{2.08^{2}}{15})^{2}}{15-1}}$

$=26.55\\\approx 27$

Compute the critical value for α = 0.05 as follows:

$t_{\alpha/2, m}=t_{0.025, 27}=\pm2.052$

*Use a t-table for the values.

Thus, the critical values of t are ± 2.052.

3 0

2 years ago