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3 years ago

8

Yesterday’s temperature in Florida was 70°F while New Jersey was -2°F. How much warmer was Florida’s temperature than NJ’s tempe

rature?

1 answer:

Marina86 [1]3 years ago

3 0

Answer:

72

Step-by-step explanation:

70-(-2)=70+2=72

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A question paper contains 10 questions, 5 marks are awarded for every correct answer and (-2) marks for every incorrect answer a

Mila [183]

Answer:

$correct = (5 \times 5) = 25 \: marks \\ incorrect = ( - 2 \times 5) = - 10 \: marks \\ unattempted = 0 \\ total = (25 - 10) \\ = 15 \: marks$

3 0

3 years ago

The ages of a random sample of five university professors are 39, 54, 61, 72, and 59. Using this information, find a 99% confide

kondor19780726 [428]

Answer:

99% confidence interval for the population standard deviation = (74.97 , 635.20).

Step-by-step explanation:

We are given that the ages of a random sample of five university professors are 39, 54, 61, 72 and 59. Also, it is provided that the ages of university professors are normally distributed.

So, firstly the pivotal quantity for 99% confidence interval for the population standard deviation is given by;

P.Q. = $\frac{(n-1)s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

where, s = sample standard deviation

$\sigma$ = population standard deviation

n = sample of university professors = 5

Also, $s^{2}$ = $\frac{\sum (X-\bar X)^{2} }{n-1}$ = 144.5

So, 99% confidence interval for population standard deviation, $\sigma$ is;

P(0.2070 < $\chi^{2} __5_-_1$ < 14.86) = 0.99 {As the table of $\chi^{2}$ at 4 degree of freedom

gives critical values of 0.2070 & 14.86}

P(0.2070 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 14.86) = 0.99

P( $\frac{ 0.2070}{(n-1)s^{2} }$ < $\frac{1 }{\sigma^{2} }$ < $\frac{ 14.86}{(n-1)s^{2} }$ ) = 0.99

P( $\frac{ (n-1)s^{2}}{14.86 }$ < $\sigma^{2}$ < $\frac{ (n-1)s^{2}}{0.2070 }$ ) = 0.99

99% confidence interval for $\sigma^{2}$ = ( $\frac{ (n-1)s^{2}}{14.86 }$ , $\frac{ (n-1)s^{2}}{0.2070 }$ )

= ( $\frac{ (5-1) \times 144.5^{2}}{14.86 }$ , $\frac{ (5-1) \times 144.5^{2}}{0.2070 }$ )

= (5620.525 , 403483.092)

99% confidence interval for $\sigma$ = ( $\sqrt{5620.525}$ , $\sqrt{403483.092}$ )

= (74.97 , 635.20)

Therefore, 99% confidence interval for the population standard deviation of the ages of all professors at the university is (74.97 , 635.20).

8 0

4 years ago

An investment of $8200 loses value at a rate of 2% per year; 7 years.<br> (growth and decrease.)

labwork [276]

Answer:

$7118.629372623104

Step-by-step explanation:

$8200 × 0.98 ^7 {{{ 7 years

/\

takes away 2%

8 0

4 years ago

Sarika’s house is 5 km from her office. She covered 4 ½ by train and the remaining by walk.

givi [52]

Answer:

1/2 km or 0.5 km

Step-by-step explanation:

In the attached file

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4 years ago

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What is LN?<br> O A 14<br> OB. 10<br> O C. 16<br> D. 7

tamaranim1 [39]

Answer:D.7

Step-by-step explanation:

6 0

4 years ago