Which expression is equivalent to the expression below?

Arun has captured many yellow-spotted salamanders. He weights each and then counts the number of yellow spots on its back

vichka [17]

Answer:

B

Step-by-step explanation:

This is a positive anotation.

Hope this helps have a good day!

7 0

3 years ago

Read 2 more answers

Determine intervals y= -x^3 + 3x^2 -2

motikmotik

Answer:

(2,2) and (0,-2)

Step-by-step explanation:

Use the website desmos for questions like graphing

3 0

3 years ago

I need help with this question

Artemon [7]

8x+5xsquared+3
tucjvkbhlnhgjfcdxfchvjgbkhnjlhkgfdgchgvhb

6 0

3 years ago

Newly purchased tires of a particular type are supposed to be filled to a pressure of 30 psi. Let m denote the true average pres

dimulka [17.4K]

Answer:

a) 48.21 %

b) 45.99 %

c) 20.88 %

d) 42.07 %

e) 50 %

Note: these values represent differences between z values and the mean

Step-by-step explanation:

The test to carry out is:

Null hypothesis H₀ is μ₀ = 30

The alternative hypothesis m ≠ 30

In which we already have the value of z for each case therefore we look directly the probability in z table and carefully take into account that we had been asked for differences from the mean (0.5)

a) z = 2.1 correspond to 0.9821 but mean value is ubicated at 0.5 then we subtract 0.9821 - 0.5 and get 0.4821 or 48.21 %

b) z = -1.75 P(m) = 0.0401 That implies the probability of m being from that point p to the end of the tail, the difference between this point and the mean so 0.5 - 0.0401 = 0.4599 or 45.99 %

c) z = -.55 P(m) = 0.2912 and this value for same reason as before is 0.5 - 0.2912 = 0.2088 or 20.88 %

d) z = 1.41 P(m) = 0.9207 0.9207 -0.5 0.4207 or 42.07 %

e) z = -5.3 P(m) = 0 meaning there is not such value in z table is too small to compute and difference to mean value will be 0.5

d) z= 1.41 P(m) =

4 0

2 years ago

Two machines are used for filling plastic bottles to a net volume of 16.0 ounces. A member of the quality engineering staff susp

aleksandrvk [35]

Answer:

Step-by-step explanation:

Hello!

The objective is to test if the two machines are filling the bottles with a net volume of 16.0 ounces or if at least one of them is different.

The parameter of interest is μ₁ - μ₂, if both machines are filling the same net volume this difference will be zero, if not it will be different.

You have one sample of ten bottles filled by each machine and the net volume of the bottles are measured, determining two variables of interest:

Sample 1 - Machine 1

X₁: Net volume of a plastic bottle filled by the machine 1.

n₁= 10

X[bar]₁= 16.02

S₁= 0.03

Sample 2 - Machine 2

X₂: Net volume of a plastic bottle filled by the machine 2

n₂= 10

X[bar]₂= 16.01

S₂= 0.03

With p-values 0.4209 (for X₁) and 0.6174 (for X₂) for the normality test and using α: 0.05, we can say that both variables of interest have a normal distribution:

X₁~N(μ₁;δ₁²)

X₂~N(μ₂;δ₂²)

Both population variances are unknown you have to conduct a homogeneity of variances test to see if they are equal (if they are you can conduct a pooled t-test) or different (if the variances are different you have to use the Welche's t-test)

H₀: δ₁²/δ₂²=1

H₁: δ₁²/δ₂²≠1

α: 0.05

$F= \frac{S^2_1}{S^2_2} * \frac{Sigma^2_1}{Sigma^2_2} ~~F_{n_1-1;n_2-1}$

Using a statistic software I've calculated the test

$F_{H_0}= 1.41$

p-value 0.6168

The p-value is greater than the significance level, so the decision is to not reject the null hypothesis then you can conclude that both population variances for the net volume filled in the plastic bottles by machines 1 and 2 are equal.

To study the difference between the population means you can use

$t= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}$

The hypotheses of interest are:

H₀: μ₁ - μ₂ = 0

H₁: μ₁ - μ₂ ≠ 0

α: 0.05

$Sa= \sqrt{\frac{(n_1-1)S^2_1+(N_2-1)S^2_2}{n_1+n_2-2}} = \sqrt{\frac{9*9.2*10^{-4}+9*6.5*10^{-4}}{10+10-2} } = 0.028= 0.03$

$t_{H_0}= \frac{(16.02-16.01)-0}{0.03*\sqrt{\frac{1}{10} +\frac{1}{10} } } = 0.149= 0.15$

The p-value for this test is: 0.882433

The p-value is greater than the significance level, the decision is to not reject the null hypothesis.

Using a significance level of 5%, there is no significant evidence to reject the null hypothesis. You can conclude that the population average of the net volume of the plastic bottles filled by machine one and by machine 2 are equal.

I hope this helps!

8 0

3 years ago