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2 years ago

Please help... How many odd perfect squares are between 5 and 211?

Mathematics

1 answer:

frez [133]2 years ago

8 0

Answer:

we conclude that the total number of perfect odd squares between 5 and 211 will be: 6

Step-by-step explanation:

Let us check by taking squares

3² = 9

5² = 25

7² = 49

9² = 81

11² = 121

13² = 169

As taking 14² = 256 would exceed 211, and 1² = 1 is smaller than 5.

Therefore, we conclude that the total number of perfect odd squares between 5 and 211 will be: 6

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If two lines have the same slope, they are _______.

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Answer:

same slope is parallel lines

Step-by-step explanation:

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3 years ago

Find the 10th term of an arithmetic sequence if the common difference is 32 and the first term is 2

zaharov [31]

Answer:

290

Step-by-step explanation:

If the first term is 2 and the common difference is 32, then the formula for a(n) is:

a(n) = 2 + 32(n-1).

Thus, the 10th term is a(10) = 2 + 32(10-1) = 2 + 32(9) = 290

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2 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

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3 years ago

Ten students share 9 candy bars equally. what fraction of a candy bars does each student get

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Each person gets 9 tenths

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2 years ago

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I need help please....

Zina [86]

Answer:

1/10

Step-by-step explanation:

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2 years ago

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