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3 years ago

Given a = -3, b=4 and c= -5, evaluate a + b + c. 04 0-6 -12

Mathematics

2 answers:

Rainbow [258]3 years ago

6 0

Answer: 0 - 4

Step-by-step explanation:

0 - 4 = -4

-3 + 4 = 1

1 - 5 = -4

White raven [17]3 years ago

5 0

Answer:

a+b+c

= -3+4+(-5)

= 1-5

= -4

So,the answer is -4.Although I don't see it among the options, but I'm sure the answer is -4

I hope I could help you. :)

Step-by-step explanation:

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Andre45 [30]

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2 years ago

Two hundred dollars is increased by 25% and the result decreased by 25%

erastova [34]

Answer:

If $200 increased by 25%, your result would be $250

If it then decreased by 25%, your result would be $187.5

Step-by-step explanation:

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5 0

2 years ago

Read 2 more answers

A student researcher compares the heights of American students and non-American students from the student body of a certain coll

OleMash [197]

Answer:

98% confidence interval for the true mean difference between the mean height of the American students and the mean height of the non-American students is [0.56 inches , 6.04 inches].

Step-by-step explanation:

We are given that a random sample of 12 American students had a mean height of 70.2 inches with a standard deviation of 2.73 inches.

A random sample of 18 non-American students had a mean height of 66.9 inches with a standard deviation of 3.13 inches.

Firstly, the Pivotal quantity for 98% confidence interval for the difference between the true means is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t__n__1-_n__2-2$

where, $\bar X_1$ = sample mean height of American students = 70.2 inches

$\bar X_2$ = sample mean height of non-American students = 66.9 inches

$s_1$ = sample standard deviation of American students = 2.73 inches

$s_2$ = sample standard deviation of non-American students = 3.13 inches

$n_1$ = sample of American students = 12

$n_2$ = sample of non-American students = 18

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(12-1)\times 2.73^{2} +(18-1)\times 3.13^{2} }{12+18-2} }$ = 2.98

<em>Here for constructing 98% confidence interval we have used Two-sample t test statistics.</em>

So, 98% confidence interval for the difference between population means ( $\mu_1-\mu_2$ ) is ;

P(-2.467 < $t_2_8$ < 2.467) = 0.98 {As the critical value of t at 28 degree

of freedom are -2.467 & 2.467 with P = 1%}

P(-2.467 < $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 2.467) = 0.98

P( $-2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ${(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}$ < $2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.98

P( $(\bar X_1-\bar X_2)-2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.98

<u>98% confidence interval for</u> ( $\mu_1-\mu_2$ ) =

[ $(\bar X_1-\bar X_2)-2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+2.467 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ]

= [ $(70.2-66.9)-2.467 \times {s_p\sqrt{\frac{1}{12} +\frac{1}{18} } }$ , $(70.2-66.9)+2.467 \times {s_p\sqrt{\frac{1}{12} +\frac{1}{18} } }$ ]

= [0.56 , 6.04]

Therefore, 98% confidence interval for the true mean difference between the mean height of the American students and the mean height of the non-American students is [0.56 inches , 6.04 inches].

3 0

3 years ago