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3 years ago

If we reject the null hypothesis, h0: ρ = 0 , what can we conclude about the population correlation coefficient?

1 answer:

5 0

If the null hypothesis, $H_0:\:\rho =0$ is rejected, then we can conclude that the correlation coefficient is significant. This means that there is enough evidence to conclude that a relationship exists between the two (or more) variables involved in the regression.

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Shirley wants to buy a skateboard for $64. She has $98 in her account. She spent $10.85 to buy stationary. She also wants to buy

klemol [59]

she had 98 but spent 10.85 so she now has:

98-10.85 = 87.15

skateboard cost 64

87.15 -64 = 23.15 is what she can spend on cookies

23.15 / 1.65 = 14.03 she can't buy more than 14 cookies

so answer is c. n ≤ 14

5 0

3 years ago

Read 2 more answers

Evaluate S(3)=10+20log4(3+1)

dimulka [17.4K]

Asssuming you mean

evaluate $10+20log_4(3+1)$

remember $log_a(a)=1$
so
$10+20log_4(3+1)$ =
$10+20log_4(4)$ =
$10+20(1)$ =
10+20=
30

4 0

3 years ago

Use long division to find the quotient Q(x) and the remainder R(x) when P(x) is divided by d(x) and express P(x) in the for

lilavasa [31]

Answer:

$P(x)=(x+9)(x^2-7x+52)+3$

Step-by-step explanation:

We are given that

$P(x)=x^3+2x^2-11x+471$

$d(x)=x+9$

We have to find the value of Q(x) and Remainder R(x).

Quotient, $Q(x)=x^2-7x+52$

Remainder, $R(x)=3$

We know that

$Dividend=Divisor\times Quotient+Remainder$

$P(x)=(x+9)(x^2-7x+52)+3$

4 0

3 years ago

A grower believes that one in five of his citrus trees are infected with the citrus red mite. How large a sample should be taken

mihalych1998 [28]

Answer:

n=6147

Step-by-step explanation:

1) Notation and definitions

$X=1$ number of citrus trees that are infected with the citrus red mite.

$n=5$ random sample taken

$\hat p=\frac{1}{5}=0.2$ estimated proportion of citrus trees that are infected with the citrus red mite.

$p$ true population proportion of citrus trees that are infected with the citrus red mite.

Me=0.01 represent the margin of error desired

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

In order to find the critical values we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical values would be given by: