What is the ratio 7 to 12 as a fraction in simplest form?

The function {(1, 4), (2, 5), (3, 6)} has the function rule y = 3x + 1.

inna [77]

False.
y= 3(1)+1= 4
y= 3(2)+1=7 not 5
y= 3(3)+1=10 not 6

4 0

4 years ago

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Find the unknown value 18 is 50% of

nasty-shy [4]

Question : 18 is 50% of what number?

Answer : 36.

How : well if you multiply 18 times 2 the answer is 36. Since 18 is half of 36, being 50%

-procklown

6 0

4 years ago

Can anyone answer this for me?

Sunny_sXe [5.5K]

Answer:

y = 2 - $x^{2}$

Step-by-step explanation:

$x^{2}$ results in a parabola (U-shape). Adding a negative in front of it flips the parabola to look like an upside-down U.

The 2 makes it shift up two decimal spots to (0,2).

8 0

3 years ago

For class trip the teachers would like to have one adult for every 10 students there are 190 students on the trip how many adult

dalvyx [7]

19 teachers

I hope this helps

have a great day

8 0

3 years ago

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An automobile manufacturer has given its van a 31.3 miles/gallon (MPG) rating. An independent testing firm has been contracted t

Fynjy0 [20]

Answer:

$z=\frac{31.1-31.3}{\frac{1.3}{\sqrt{140}}}=-1.82$

The p value for this case would be given by:

$p_v =2*P(z$

For this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly different from 31.3 MPG

Step-by-step explanation:

Information given

$\bar X=31.1$ represent the sample mean

$\sigma=1.3$ represent the population standard deviation

$n=140$ sample size

$\mu_o =31.3$ represent the value that we want to test

$\alpha=0.02$ 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 is equal to 31.3 MPG, the system of hypothesis would be:

Null hypothesis: $\mu =31.3$

Alternative hypothesis: $\mu \neq 31.3$

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:

$z=\frac{31.1-31.3}{\frac{1.3}{\sqrt{140}}}=-1.82$

The p value for this case would be given by:

$p_v =2*P(z$

For this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly different from 31.3 MPG

4 0

3 years ago