12 is at most a number decreased by 7

Does the data suggest a linear, quadratic, or exponential model?

oksano4ka [1.4K]

Exponential model because it's increasing rapidly.

4 0

3 years ago

Carlos started a business with 8 customers. Each month, his business increased by 4 customers. He would like 52 customers a mont

maxonik [38]

Answer: 11 months

Step-by-step explanation:

Let the number of months that it will take him to reach his goal be represented by x.

Based on the information given in the question, this can be formed into an equation as:

8 + 4x = 52

4x = 52 - 8

4x = 44

x = 44/4

x = 11

It will take 11 months to reach the goal

5 0

3 years ago

A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro

-Dominant- [34]

Answer:

$t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07$

The degrees of freedom are given by:

$df =n-1= 32-1=31$

Since is a two-sided test the p value would be:

$p_v =2*P(t_{31}>3.07)=0.0044$

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example ( $\alpha=0.01,0.05, 0.1, 0.15$ ).

Step-by-step explanation:

Data given and notation

$\bar X=1.6$ represent the sample mean

$s=0.46$ represent the sample deviation

$n=32$ sample size

$\mu_o =1.35$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:

Null hypothesis: $\mu =1.35$

Alternative hypothesis: $\mu \neq 1.35$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07$

P-value

The degrees of freedom are given by:

$df =n-1= 32-1=31$

Since is a two-sided test the p value would be:

$p_v =2*P(t_{31}>3.07)=0.0044$

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example ( $\alpha=0.01,0.05, 0.1, 0.15$ ).

5 0

3 years ago

Suppose a normal distribution has a mean of 38 and a standard deviation of 2. What is the probability that a data value is betwe

Dimas [21]

Answer:

P ( 37 < x < 41) = P(-0.5 < Z < 1.5) = 0.6247

Step-by-step explanation:

We know mean u = 38 standard dev. s = 2

We want P ( 37 < x < 41)

so

P( (37 - 38) / 2 < Z) = P(-0.5 < Z)

P( Z < (41 - 38)/2 ) = P( Z < 1.5)

Find P(Z < -0.5) = 0.3085

Find P(Z > 1.5) = 0.0668

so P(-0.5 < Z < 1.5) = 1 - P(Z < -0.5) - P(Z > 1.5)

P(-0.5 < Z < 1.5) = 1 - 0.3085 - 0.0668

P(-0.5 < Z < 1.5) = 0.6247

P ( 37 < x < 41) = P(-0.5 < Z < 1.5) = 0.6247

3 0

3 years ago

Right triangle similarity. what is the value of a?

kramer

Answer:

I'd say that a is 6 2/3 units long

Step-by-step explanation:

4 0

3 years ago