Your cell phone company started a rewards club. For every three texts sent, you get 15

Which is the solution set of the inequality 15y-9<36

m_a_m_a [10]

15y - 9 <36.
Add 9 to each side.

15y- 9 (+9) < 36 + 9.
15y < 45.
y < 3.
Hope it helps! :)

7 0

3 years ago

A standard American Eskimo dog has a mean weight of 30 pounds with a standard deviation of 2 pounds. Assuming the weights of sta

nalin [4]

Answer:

Option 2 - Approximately 24–36 pounds

Step-by-step explanation:

Given : A standard American Eskimo dog has a mean weight of 30 pounds with a standard deviation of 2 pounds. Assuming the weights of standard Eskimo dogs are normally distributed.

To find : What range of weights would 99.7% of the dogs have?

Solution :

The range of 99.7% will lie between the mean ± 3 standard deviations.

We have given,

Mean weight of Eskimo dogs is $\mu=30$

Standard deviation of Eskimo dogs is $\sigma=2$

The range of weights would 99.7% of the dogs have,

$R=\mu\pm3\sigma$

$R=30\pm3(2)$

$R=30\pm6$

$R=30+6,30-6$

$R=36,24$

Therefore, The range is approximately, 24 - 36 pounds.

So, Option 2 is correct.

5 0

4 years ago

The graph below shows the heights and arm spans of students in a classroom. Height and Arm Span of Students Based on the graph,

xxMikexx [17]

The best prediction of the height of a student with an arm span of 143 cm is 143 cm

Step-by-step explanation:

The missing graph is attached

The graph below shows the heights and arm spans of students in

a classroom

The x-axis represents the height in cm
The y-axis represents the arm span in cm

We need to find the height of a student with an arm span of 143 cm

∵ The graph does not contain the value of the given arm span

∴ We will use the line best fit (sold line in the attached graph) to find

the best predication of the height

∵ The line passes through points (150 , 150) and (160 , 160)

- The x-coordinates are equal the y-coordinate

∴ The equation of the line is y = x

∵ The arm span of a student is 143 cm

∵ y represents the arm span of a student

∴ y = 143

- Substitute the value of y in the equation of the line best fit

∵ y = x

∴ 143 = x

∵ x represents the height of a student

∴ The height of the student is 143 cm

The best prediction of the height of a student with an arm span of 143 cm is 143 cm

Learn more:

You can learn more about line best fit in brainly.com/question/8955867

#LearnwithBrainly

3 0

3 years ago

Read 2 more answers

The breaking strengths of cables produced by a certain manufacturer have a mean, u, of 1750 pounds, and a standard deviation of

AlladinOne [14]

Answer:

We accept the null hypothesis that the breaking strength mean is less and equal to 1750 pounds and has not increased.

Step-by-step explanation:

The null and alternative hypotheses are stated as

H0: u ≥ 1750 i.e the mean is less and equal to 1750

against the claim

Ha: u > 1750 ( one tailed test) the mean is greater than 1750

Sample mean = x`= 1754

Population mean = u = 1750

Population deviation= σ = 65 pounds

Sample size= n = 100

Applying the Z test

z= x`- u / σ/ √n

z= 1754- 1750 / 65/ √100

z= 4/6.5

z= 0.6154

The significance level alpha = 0.1

The z - value at 0.1 for one tailed test is ± 1.28

The critical value is z > z∝.

so

0.6154 is < 1.28

We accept the null hypothesis that the breaking strength mean is less and equal to 1750 pounds and has not increased.

8 0

3 years ago

A rare form of malignant tumor occurs in 11 children in a million, so its probability is 0.000011. Four cases of this tumor occ

Shkiper50 [21]

Answer:

a) The mean number of cases is 0.14608 cases.

b) The probability that the number of cases is exactly 0 or 1 is 0.990.

c) The probability of more than one case is 0.010

d) No, because the probability of more than one case is very small

Step-by-step explanation:

We can model this problem with a Poisson distribution, with parameter:

$\lambda=r*t=0.000011*13,280=0.14608$

a) The mean amount of cases is equal to the parameter λ=0.14608.

b) The probability of having 0 or 1 cases is:

$P(k=0)=\frac{\lambda^0 e^{-\lambda}}{0!}=\frac{1*0.864}{1} =0.864\\\\ P(k=1)=\frac{\lambda^1 e^{-\lambda}}{0!}=\frac{0.14608*0.864}{1} =0.126\\\\P(k\leq1)=0.864+0.126=0.990$

c) The probability of more than one case is:

$P(k>1)=1-P(k\leq 1)=1-0.990=0.010$

d) The cluster of 4 cases can not be due to pure chance, as it is a very high proportion of cases according to the average rate. Just having more than one case has a probability of 1%.

7 0

3 years ago