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4 years ago

11

Four-ninths of those who rode the giant's gyro at the fair were delirious. All the rest were dazed. What fraction of those who r

ode the giant gyro was dazed

1 answer:

Paraphin [41]4 years ago

8 0

Answer:

5/9's of those who rode the giant gyro were dazed

Step-by-step explanation:

if 4/9's were delirious then the other 5/9's were dazed

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Nora uses 2 ft of ribbon to make a bow she has 8 yards of ribbon . how many bows can she make

7nadin3 [17]

She can make 12 bows because there are 3 feet in one yard so 8*3 is 24. then you divide 24 by 2 and you get 12

5 0

4 years ago

Read 2 more answers

64 - 9p for p=4,9,-3

V125BC [204]

64-9(4)

64-36=28

64-9(9)

64-81=-17

64-9(-3)

64+27=91

3 0

4 years ago

Read 2 more answers

A random sample of 10 parking meters in a resort community showed the following incomes for a day. Assume the incomes are normal

GenaCL600 [577]

Answer:

A 95% confidence interval for the true mean is [$3.39, $6.01].

Step-by-step explanation:

We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;

Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean income = $\frac{\sum X}{n}$ = $4.70

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = $1.83

n = sample of parking meters = 10

$\mu$ = population mean

<em>Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.</em>

<u>So, 95% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-2.262 < $t_9$ < 2.262) = 0.95 {As the critical value of t at 9 degrees of

freedom are -2.262 & 2.262 with P = 2.5%}

P(-2.262 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.262) = 0.95

P( $-2.262 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu$ < $2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

<u>95% confidence interval for</u> $\mu$ = [ $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $4.70-2.262 \times {\frac{1.83}{\sqrt{10} } }$ , $4.70+ 2.262 \times {\frac{1.83}{\sqrt{10} } }$ ]

= [$3.39, $6.01]

Therefore, a 95% confidence interval for the true mean is [$3.39, $6.01].

The interpretation of the above result is that we are 95% confident that the true mean will lie between incomes of $3.39 and $6.01.

Also, the margin of error = $2.262 \times {\frac{s}{\sqrt{n} } }$

= $2.262 \times {\frac{1.83}{\sqrt{10} } }$ = <u>1.31</u>

4 0

3 years ago

Or an angle θ with the point (−20, −21) on its terminating side, what is the value of cosine?

kherson [118]

Cos(θ) = -20/√((-20)^2 +(-21)^2)
.. = -20/-√841 . . . . . . . . . . . . . . . . . θ is a 4th-quadrant angle, so cos(θ) > 0
.. = 20/29

The value of the cosine is 20/29.

3 0

3 years ago

Read 2 more answers

A) The y-intercept of the line with the largest slope

lord [1]

Answer:

a) the largest y-intercept is 1

b) 13.5

c) y=x-3

d) 8

Step-by-step explanation:

the question was worded very strangely and I will answer the way I precieved it.

for question d, I simply considered the question to be asking what is X when y is 17 and since the equation of that line is 2x-1

5 0

3 years ago