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2 years ago

50 POINTS!!!! PLEASE HELP WILL GIVE BRAINLIEST!!! :3

Mathematics

1 answer:

ratelena [41]2 years ago

5 0

Answer:

The student solved it wrong, and the graph also shows it wrong (you can pick the answer yourself)

Step-by-step explanation:

6(x-0,5) > 3
6x - 3 > 3
6x > 6
x> 1

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The angles of a triangle are in a ratio 1:3:8. What are the angles?<br>

xxTIMURxx [149]

Answer:

15° , 45°, 120°

Step-by-step explanation:

sum the parts of the ratio, 1 + 3 + 8 = 12 parts

the sum of the 3 angles in a triangle = 180° , then

180° ÷ 12 = 15° ← value of 1 part of the ratio, then

3 parts = 3 × 15° = 45°

8 parts = 8 × 15° = 120°

the 3 angles are 15° , 45° , 120°

6 0

3 years ago

The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out t

yulyashka [42]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 0.5 \ in$

Generally the Null hypothesis is $H_o : \mu = 0. 5 \ in$

The Alternative hypothesis is $H_a : \mu \ne 0.5 \ in$

Considering the parameter given for part a

The sample size is n = 15

The test statistics is t = 1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = 2.145$

We are making use of this $t_{\frac{\alpha }{2} }$ because it is a one-tail test

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that $t < t_{\frac{\alpha }{2} }$ so the null hypothesis would not be rejected

Considering the parameter given for part b

The sample size is n = 15

The test statistics is t = -1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = -2.145$

Looking at the value of t and $t_{\frac{\alpha}{2} ,df }$ the we see that t does not lie in the area covered by $t_{\frac{\alpha}{2} , df }$ (i.e the area from -2.145 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part c

The sample size is n = 26

The test statistics is t = -2.55

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = 2.787$

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that t does not lie in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part d

The sample size is n = 26

The test statistics is t = -3.95

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = -2.787$

Looking at the value of t and $t_{\frac{\alpha}{2} }$ the we see that t lies in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we reject the null hypothesis

6 0

3 years ago

Can you pls help me

SashulF [63]

I answered the last post, the third option is correct

7 0

3 years ago

A survey of 1,108 employees at a software company finds that 621 employees take a bus to work and 445 employees take a train to

anastassius [24]

We know 445 employees take the train, and that 321 of these exclusively take the train. So 445 - 321 = 124 take both the train and bus.

Now, if $B$ is the set of employees that take the bus and $T$ the set of employees that take the train, then

$n(B\cup T)=n(B)+n(T)-n(B\cap T)=621+445-124=942$

where $n(A)$ is the number of employees belonging to a general set $A$ .

So the probability that an employee takes either the bus or train is

$\dfrac{n(B\cup T)}{1108}=\dfrac{942}{1108}\approx85\%$

4 0

4 years ago

City A: 3% City B: 4.1% City C: 2.9% City D: 3.2% Tom wants to buy a sweater for $30. In witch city will it be lest expensive?

guapka [62]

The answer is C 2.9%

6 0

4 years ago

Read 2 more answers