Ask question

Ask question

All categories

3 years ago

7

in the state of manie,the ratio of moose to people is 5 to 2.If a certain county in Maine has a population of 1,120 people, how

many moose would you expect to live in the country?

1 answer:

viva [34]3 years ago

3 0

Where is the state of "manie"?

You might be interested in

yulyashka [42]

Answer:

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Explain how the graph of a quadratic function relates to

igomit [66]

The equation for a quadratic is as follows:
y = ax^2+bx+c

The first variable “a” determines whether the parabola opens up or down
The squared allows the equation to look like the parabola swoop when graphed.
The c term is the constant term

5 0

3 years ago

Read 2 more answers

Ms. Ramirez has a class of 15 students. She can spend $18 on each student to buy math supplies for the year. She first buys all

Aloiza [94]

Answer:7.775

Step-by-step explanation:

So u do 130.05 ÷18 =7.225

Now u do 15 - 7.225 = 7.775

7.775 is the money she can still spend for them.

Also u could do 15 x 18 = 270

270 ÷ 2 = 135

8 0

3 years ago

.A variety of stores offer loyalty programs. Participating shoppers swipe a bar-coded tag at the register when checking out and

Leni [432]

Answer:

a) Null hypothesis: $\mu \leq 120$

Alternative hypothesis: $\mu > 120$

b) $t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236$

The degrees of freedom are given by:

$df = n-1 = 80-1=79$

The p value for this case taking in count the alternative hypothesis would be:

$p_v =P(t_{79}>2.236)=0.0141$

Step-by-step explanation:

Information given

$\bar X=130$ represent the sample mean for the amount spent each shopper

$s=40$ represent the sample standard deviation

$n=80$ sample size

$\mu_o =120$ represent the value to verify

t would represent the statistic

$p_v$ represent the p value f

Part a

We want to verify if the shoppers participating in the loyalty program spent more on average than typical shoppers, the system of hypothesis would be:

Null hypothesis: $\mu \leq 120$

Alternative hypothesis: $\mu > 120$

The statistic for this case would be given by:

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

Replacing the info given we got:

$t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236$

The degrees of freedom are given by:

$df = n-1 = 80-1=79$

The p value for this case taking in count the alternative hypothesis would be:

$p_v =P(t_{79}>2.236)=0.0141$

5 0

3 years ago

Write Five hundred thirty-eight thousandths in numbers

Vedmedyk [2.9K]

.538 should be correct

4 0

3 years ago