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3 years ago

3) Keith took a ferry to a nearby island that lasted for 82 minutes. Although he forgot to check the

Mathematics

1 answer:

adell [148]3 years ago

6 0

Answer:

Step-by-step explanation:

so if he arrived at 11:22 AM.....and it took him 82 minutes....

82 minutes = 1 hr and 22 minutes

then he departed at :

11:22 - 1 hr and 22 minutes = 10:00 AM <==

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What are the coordinates of the focus of the parabola shown below?

soldi70 [24.7K]

The coordinates are (-4,0),(0,-1)

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3 years ago

Who can give me this answer

-BARSIC- [3]

The believe the answer is 3/5

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3 years ago

I'm confused What is the answer?

forsale [732]

If c > 0, then f(x - c) is a shift of f(x) by c units to the right, and f(x + c) is a shift by c units to the left.

If d > 0, then f(x) - d is a shift by d units downward, and f(x) + d is a shift by d units upward.

Let g(x) = x. Then f(x) = g(x + a) - b = (x + a) - b. So to get g(x), we translate f(x) to the left by a units, and down by b units.

Note that we can also interpret the translation as

• a shift upward of a - b units, since

(x + a) - b = x + (a - b)

• a shift b units to the right and a units upward, since

(x + a) - b = x + (a - b) = x + (- b + a) = (x - b) + a.

6 0

3 years ago

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

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.

So, 95% confidence interval for the population mean, $\mu$ is ;

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

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

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

= [$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 = $2.262 \times {\frac{s}{\sqrt{n} } }$

= $2.262 \times {\frac{1.83}{\sqrt{10} } }$ = 1.31

4 0

3 years ago

3) If a = -19, d = 4 then find t2

geniusboy [140]

Answer:

-15.

Step-by-step explanation:

Arithmetic sequence:-

t2 = a + d

= -19 + 4

= -15.

8 0

3 years ago