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3 years ago

What is the exponential form of the following expression? (-5)(-5)(-5) . c . c .c

Mathematics

2 answers:

andrew-mc [135]3 years ago

7 0

Answer:

$(-5)^3 * c^3$

Step-by-step explanation:

Given

$(-5)(-5)(-5) * c * c * c$

Required

The exponent form

We have:

$(-5)(-5)(-5) * c * c * c$

Apply the following law of indices;

$x * x * x ...... = x^n$

So, the expression becomes:

$(-5)^3 * c^3$

Zarrin [17]3 years ago

7 0

Answer:

its c

Step-by-step explanation:

Did it on edge

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Write the equation in it's simplest form or as

allsm [11]

9514 1404 393

Answer:

2∛7

Step-by-step explanation:

Putting the expression in simplest form will factor cubes from under the radical.

$\displaystyle\sqrt[3]{56}=\sqrt[3]{2^3\cdot7}=\boxed{2\sqrt[3]{7}}$

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3 years ago

A farmer is planting a straight row of crops with 3 feet between each plant. If 23 are planted, what is the distance between the

densk [106]

Number of plants planted = 23
with a distance of = 3 m
total distance between the first and the last plant = 22x3 ( because starting from first plant we will not take 23)
ans. = 66m

5 0

3 years ago

The graph of a linear function passes through the point (2,5) and a slope of 3

Delicious77 [7]

Answer:

$y=3x-1$

Step-by-step explanation:

The point slope form of a line is $(y-y_1)=m(x-x_1)$ where $x_1=2\\y_1=5$ . We write

$(y-5)=3(x-2)$

Convert point slope form to slope intercept form by simplifying and rearranging.

$y-5=3(x-2)\\y-5=3x-6\\y=3x-1$

8 0

3 years ago

A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. Assume that the distr

Marianna [84]

Answer:

(a) Decision rule for 0.01 significance level is that we will reject our null hypothesis if the test statistics does not lie between t = -2.651 and t = 2.651.

(b) The value of t test statistics is 1.890.

(c) We conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) P-value of the test statistics is 0.0662.

Step-by-step explanation:

We are given that a recent study focused on the number of times men and women who live alone buy take-out dinner in a month.

Also, following information is given below;

Statistic : Men Women

The sample mean : 24.51 22.69

Sample standard deviation : 4.48 3.86

Sample size : 35 40

Let $\mu_1$ = mean number of times men order take-out dinners in a month.

$\mu_2$ = mean number of times women order take-out dinners in a month

(a) So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2$ = 0 {means that there is no difference in the mean number of times men and women order take-out dinners in a month}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2\neq$ 0 {means that there is difference in the mean number of times men and women order take-out dinners in a month}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_-_n_2_-_2$

where, $\bar X_1$ = sample mean for men = 24.51

$\bar X_2$ = sample mean for women = 22.69

$s_1$ = sample standard deviation for men = 4.48

$s_2$ = sample standard deviation for women = 3.86

$n_1$ = sample of men = 35

$n_2$ = sample of women = 40

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(35-1)\times 4.48^{2}+(40-1)\times 3.86^{2} }{35+40-2} }$ = 4.16

So, test statistics = $\frac{(24.51-22.69)-(0)}{4.16 \sqrt{\frac{1}{35}+\frac{1}{40} } }$ ~ $t_7_3$

= 1.890

(b) The value of t test statistics is 1.890.

(c) Now, at 0.01 significance level the t table gives critical values of -2.651 and 2.651 at 73 degree of freedom for two-tailed test.

Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) Now, the P-value of the test statistics is given by;

P-value = P( $t_7_3$ > 1.89) = 0.0331

So, P-value for two tailed test is = 2 $\times$ 0.0331 = 0.0662

4 0

3 years ago

A test has 50 questions. each right answer is worth 2 points; each wrong answer deducts 0.5 points; blank answers are not counte

mestny [16]

ANSWER: There are 2 blank answers

EXPLANATION

Let

The number of right answers be ‘r’

The number of wrong answers be ‘w’

The number of blank answers be ‘b’

r + w + b = 50

This means r + w ≤ 50

Then we know,

Right answers = 2 marks

Wrong answers = -0.5 mark

Blank Answers = 0 marks

2r – 0.5w = 88.5

2r = 88.5 + 0.5w … (Equation I)

Since the score as .5, we know that there is at least one wrong answer, and the number of wrong answers is an odd number.

Since the score is 88.5, and each right answer gives 2 marks

There are at more than 44 (i.e. 88/2) right answers

Since r + w ≤ 50, and possible values of w are odd numbers

If r = 45, Possible values of w are 1, 3, and 5

If r = 46, Possible values of w are 1, 3

If r = 47, Possible values of w are 1, 3

If r = 48, The only possible value of w is 1

If r = 49,The only possible value of w is 1

Since 2r = 88.5 + 0.5w (Equation I)

We test for possible values:

If r = 45

2r = 88.5 + 0.5w

2(45) = 88.5 + 0.5w

90 = 88.5 + 0.5w

0.5w = 90 – 88.5

0.5w = 1.5

w = 3

So,

If there are 45 right answers

There are 3 wrong answers

r + w + b = 50

45 + 3 + b = 50

48 + b = 50

b = 50 – 48

b = 2

Then, there are 2 blank answers.

If r = 46

2r = 88.5 + 0.5w

2(46) = 88.5 + 0.5w

92 = 88.5 + 0.5w

0.5w = 92 – 88.5

0.5w = 3.5

w = 7

So,

If there are 46 right answers

There are 7 wrong answers

We know that r + w ≤ 50

46 + 7 = 53

So 46 and higher numbers are not possible solutions.

The only possible solution is:

There are 45 right answers

There are 3 wrong answers

There are 2 blank answers

6 0

3 years ago