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3 years ago

Can someone help me with this?

Mathematics

2 answers:

Annette [7]3 years ago

7 0

I think b I’m not sure

kherson [118]3 years ago

6 0

B is what i think is right

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Specifications call for the wall thickness of two-liter polycarbonate bottles to average 4.0 mils. A quality control engineer sa

bija089 [108]

Answer:

The null hypothesis μ=4.0 could not be rejected.

Step-by-step explanation:

In this problem we have to perform a hypothesis test of the mean, with s.d. of the population unknown.

The sample is: [3.999, 4.037, 4.116, 4.063, 3.969, 3.955, 4.091]

This sample has a mean of 4.033 and a standard deviation of 0.061.

The null and alternative hypothesis are:

$H_0: \mu = 4.0\\\\H_1: \mu \neq 4.0$

We assume a significance level of 0.05.

The test statistic for this test is:

$t=\frac{M-\mu}{\sigma}=\frac{4.033-4.0}{0.061}=0.536$

We have a sample size n=7, so the degrees of freedom are 7-1=6.

For a two-tailed test, t=0.536 and df=6, the P-value can be look up in a t-table.

The P-value is 0.61. Is greater than the significance level, so the effect is not significant and the null hypothesis can't be rejected.

8 0

4 years ago

90 + 90 ÷ 90 = _________

Karolina [17]

Answer:

Step-by-step explanation:

this is correct answer

6 0

3 years ago

A ABC is isosceles. Find AC. B 3x + 3 20 + 8 A A C

Sunny_sXe [5.5K]

Ansfdfgh gggghjtyuhjrswzargfyf

Step-by-step explanation:

4 0

3 years ago

Use the quadratic formula to solve the equation. x ^2 – 7 x –6 = 0 (Show work)

Vika [28.1K]

Given a quadratic equation

$ax^2+bx+c=0$

the quadratic formula states that the solutions are

$x_{1,2}=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

Plug your values to get

$x_{1,2}=\dfrac{-(-7)\pm\sqrt{(-7)^2-4\cdot 1 \cdot (-6)}}{2\cdot 1}$

Which evaluates to

$\dfrac{-(-7)\pm\sqrt{(-7)^2-4\cdot 1 \cdot (-6)}}{2\cdot 1} = \dfrac{7\pm\sqrt{73}}{2}$

So, the two solutions are

$x_1 = \dfrac{7+\sqrt{73}}{2},\quad x_2 = \dfrac{7-\sqrt{73}}{2}$

5 0

3 years ago

Which of the following vectors are orthogonal to (-1,3)?

snow_tiger [21]

Any vector that is a multiple of (3, 1) will be orthogonal. These include
.. A (-6, -2)
.. D (3, 1)

_____
The dot-product of these with (-1, 3) is zero:
(-6*-1 +-2*3) = 6 -6 = 0
(3*-1 +1*3) = -3 +3 = 0

You can make a vector orthogonal to a 2-D vector by swapping the coordinates and negating one of them. When you swap the elements of (-1, 3) you get (3, -1). It is usually convenient to negate the one that is already negative, so that would give you (3, 1) as the orthogonal vector.

7 0

3 years ago