(14+42-6)/10+7^2 order of operation

An automobile manufacturer has given its car a 46.7 miles/gallon (MPG) rating. An independent testing firm has been contracted t

erastova [34]

Answer:

$z=\frac{46.5-46.7}{\frac{1.1}{\sqrt{150}}}=-2.23$

The p value would be given by:

$p_v =2*P(z$

For this case since th p value is lower than the significance level of0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case is significantly different from 46.7 MPG

Step-by-step explanation:

Information given

$\bar X=46.5$ represent the mean

$\sigma=1.1$ represent the population standard deviation

$n=150$ sample size

$\mu_o =46.7$ represent the value to verify

$\alpha=0.05$ represent the significance level for the hypothesis test.

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true mean for this case is 46.7, the system of hypothesis would be:

Null hypothesis: $\mu = 46.7$

Alternative hypothesis: $\mu \neq 46.7$

Since we know the population deviation the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

Replacing we got:

$z=\frac{46.5-46.7}{\frac{1.1}{\sqrt{150}}}=-2.23$

The p value would be given by:

$p_v =2*P(z$

For this case since th p value is lower than the significance level of0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case is significantly different from 46.7 MPG

7 0

2 years ago

HURRY PLZZ!!! THIS IS 90 POINTS!!! I WILL MARK BRANLYEST!!!! 2. A ladder is leaning against the outside wall of a building. The

Marina86 [1]

Answer & Step-by-step explanation:

7 0

2 years ago

Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai

marta [7]

$f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1$

Lets check with every option

(a) [-4,-3]

We plug in -4 for x and -3 for x

$f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55$

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3 for x and -2 for x

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.

(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0 for x and 1 for x

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.

(f) [1,2]

We plug in 1 for x and 2 for x

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

$f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]

8 0

2 years ago

Read 2 more answers

How much sevens does it take to make 32?

Ksju [112]

Divide 7 by 32 and go from there

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2 years ago

Joann can run the first 50 meters of the 200-meter race in 6.3 seconds. If she can maintain the same speed for the whole rece, h

ioda

Answer:

Step-by-step explanation:

shes fastr

8 0

2 years ago