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3 years ago

Ali has a quarters, d dimes, and n nickels. He gets four more quarters. He gives away three

Mathematics

1 answer:

viva [34]3 years ago

4 0

Answer:

(q + 4) + (d - 3) + n

Step-by-step explanation:

You have to add all of them.

You have to add q, d, and n.

so q + d + n

and then you add four more q's so

(q + 4) + d + n

and then you lost 3 dimes so

(q + 4) + (d - 3) + n

and nothing happened to n so that is the answer.

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Swimming Pool On a certain hot summer's day, 630 people used the public swimming pool. The

crimeas [40]

$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the linear equation in two variables.

So we have to form two equations to solve this.

Let the number of children be 'x' and number of adults be 'y'.

hence, we get as,

x+y = 630 ..... (1)

and also, to their total cost spending, we get as,

1.5x + 2.25y = 1170 ...... (2)

solving (1) and (2) we get as,

Number of children were = 330

and Number of Adults were = 300

(Solved in the attachment)

4 0

4 years ago

If a hurricane was headed your way, would you evacuate? The headline of a press release states, "Thirty-four Percent of People o

IgorC [24]

Answer:

The 98% confidence interval would be given by (0.326;0.354)

Step-by-step explanation:

1) Notation and definitions

$X=63$ number of people who live within 20 miles of the coast in high hurricane risk counties of eight southern states

$n=6138$ random sample taken

$\hat p=0.34$ estimated proportion of people who live within 20 miles of the coast in high hurricane risk counties of eight southern states

$p$ true population proportion of people who live within 20 miles of the coast in high hurricane risk counties of eight southern states

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

2) Confidence interval

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 98% of confidence, our significance level would be given by $\alpha=1-0.98=0.02$ and $\alpha/2 =0.01$ . And the critical value would be given by: