F(x)=5x^2+5x-30 <br> if anyone can help me that would be great

A map scale is give as 1:10000

Fynjy0 [20]

Answer:

2km

Step-by-step explanation:

6 0

3 years ago

company with a large fleet of cars hopes to keep gasoline costs down and sets a goal of attaining a fleet average of at least 26

podryga [215]

Answer:

$t=\frac{25.02-26}{\frac{4.83}{\sqrt{50}}}=-1.435$

$p_v =P(t_{(49)}$

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its significant lower than 26 so then the specification is satisfied.

Step-by-step explanation:

Data given and notation

$\bar X=25.02$ represent the sample mean

$s=4.83$ represent the sample standard deviation

$n=50$ sample size

$\mu_o =26$ represent the value that we want to test

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

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is at least 26 mpg, the system of hypothesis would be:

Null hypothesis: $\mu \geq 26$

Alternative hypothesis: $\mu < 26$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{25.02-26}{\frac{4.83}{\sqrt{50}}}=-1.435$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=50-1=49$

Since is a one sided test the p value would be:

$p_v =P(t_{(49)}$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its significant lower than 26 so then the specification is satisfied.

3 0

3 years ago

Can someone help me with this please?

AVprozaik [17]

Answer:

y=70°

z= 60°

w= 70°

Step-by-step explanation:

Since the two triangles are similar, we know that all of the corresponding angles must be the same. On the triangle on the left, we can see that the angle formed between sides AB and AC is 60°. This means that the same angle is formed at Z, which is in the same location on the triangle to the right. From there, we can find the missing angle, W, from the difference of 180°.

180°-50°-60°=70°.

The same way we found Z, we can also find Y, since it has to be the same as W.

8 0

3 years ago

Write an equation of the line containing the given point and parallel to the given line.

Dafna11 [192]

Answer:

y = 5/9x - 37/3

Step-by-step explanation:

5x - 9y = 8

-9y = -5x + 8

y = 5/9x - 8/9

When a line is parallel, the slope remains the same.

y = 5/9x + b

-9 = 5/9(6) + b

-9 = 10/3 + b

-37/3 = b

6 0

2 years ago

Awarding road contracts. A state Department of Transportation (DOT) recently claimed that each of five bidders received equal co

Sveta_85 [38]

Answer:

a) Hence, the probability of this event occurring if, in fact, the DOT's claim is true

= 3/10 = 0.3

Step-by-step explanation:

a) The concept applied here is combinatorics ; nCr = n!/(n - r)!r!

we are told that out of 5bidders, Three of the bidders were large construction conglomerates

and two were small specialty contractors.

selecting two recipients with tree of the bidders were large construction conglomerates 3C2

= 3!/(3-2)!2!

= 3

Similarly, selecting the two small casualty contractor from the total of 5bidders = 5C2

= 5!/(5-2)!2!

= 10

Hence, the probability of this event occurring if, in fact, the DOT's claim is true

= 3/10 = 0.3

b) if DOTs claim is not true ( probability of failure) = 1 - 0.3

= 0.7 , this value implies that the probability 0.3 is largely compatible and a compliment of the 0.7 and as such it is consistent with the DOTs claim that the selection was random.

7 0

3 years ago

F(x)=5x^2+5x-30 if anyone can help me that would be great