What is the value of 14:45

Travka [436]

Answer: class B

Explanation:
Normal distributions are bell-shaped and symmetric about the mean. So, if the data is scattered around a lot, like in Class A & B, it would not be normal.

7 0

3 years ago

What method would be best to use 53=4(x-3)^2-11

Veronika [31]

Answer:

x = 7 , -1

Step-by-step explanation:

<u>SOLUTION :-</u>

$4(x-3)^2-11 = 53$

Add 11 to both the sides.

$=> 4(x-3)^2-11+11=53+11$

$=> 4(x-3)^2 = 64$

Divide both the sides by 4.

$=> \frac{4(x-3)^2}{4} = \frac{64}{4}$

$=> (x-3)^2 = 16$

Root square both the sides.

$=> \sqrt{(x-3)^2} = \sqrt{16}$

$=> x-3 = +4 \; or -4$

Here , x will have two values -

1) $x-3 = 4$

$=> x = 4 + 3 = 7$

2) $x - 3 = -4$

$=> x = -4 + 3 = -1$

<u>VERIFICATION :-</u>

When x = 7 ,

$4(x-3)^2 - 11 = 4(7 - 3)^2 - 11$

$= 4 \times 4^2 - 11$

$= 4 \times 16 - 11$

$= 64 - 11$

$= 53$

When x = -1 ,

$4(x-3)^2 - 11 = 4(-1 - 3)^2 - 11$

$= 4 \times (-4)^2 - 11$

$= 4 \times 16 - 11$

$= 64 - 11$

$= 53$

7 0

3 years ago

PLEASE HELP! WAYYY OVERDUE!!!

In-s [12.5K]

Answers:

Red = negative skew
Yellow = symmetric
Blue = positive skew

================================================

Explanation:

A symmetric distribution is where we have the left half be a mirror copy of the right half. In other words, we have symmetry going on.

For the yellow team, note the distance between Q1 and the median is 1 unit (47-46 = 1), and so is the distance from the median to Q3 (48-47 = 1). So the box portion has perfect symmetry. You should find that the whiskers are also perfectly symmetric as well using a similar technique of finding the distances and comparing them.

Overall, the entire box-and-whisker plot for the yellow team is perfectly symmetric.

---------------

If a distribution isn't symmetric, then it's skewed in some way. If the right side is pulled longer than the left, then we consider it "skewed to the right" aka "positively skewed". So that applies to the blue team. The red team has negative skew because its left tail is longer than the right tail.

6 0

3 years ago

Land's Bend sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the intern

Naya [18.7K]

Answer:

80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases ( $\mu_1-\mu_2$ ) is [-9.132 , 23.332].

Step-by-step explanation:

We are given that a random sample of 7 sales receipts for mail-order sales results in a mean sale amount of $81.70 with a standard deviation of $18.75.

A random sample of 11 sales receipts for internet sales results in a mean sale amount of $74.60 with a standard deviation of $28.25.

Firstly, the Pivotal quantity for 80% confidence interval for the difference between population 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 sales receipts for mail-order sales = $81.70

$\bar X_2$ = sample mean sales receipts for internet sales = $74.60

$s_1$ = sample standard deviation for mail-order sales = $18.75

$s_2$ = sample standard deviation for internet sales = $28.25

$n_1$ = size of sales receipts for mail-order sales = 7

$n_2$ = size of sales receipts for internet sales = 11

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(7-1)\times 18.75^{2} +(11-1)\times 28.25^{2} }{7+11-2} }$ = 25.11

<em>Here for constructing 80% confidence interval we have used Two-sample t test statistics as we don't know about population standard deviations.</em>

<em />

So, 80% confidence interval for the difference between population means, ( $\mu_1-\mu_2$ ) is ;

P(-1.337 < $t_1_6$ < 1.337) = 0.80 {As the critical value of t at 16 degree

of freedom are -1.337 & 1.337 with P = 10%}

P(-1.337 < $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 1.337) = 0.80

P( $-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ${(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}$ < $1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.80

P( $(\bar X_1-\bar X_2)-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.80

<u>80% confidence interval for</u> ( $\mu_1-\mu_2$ ) =

[ $(\bar X_1-\bar X_2)-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ]

= [ $(81.70-74.60)-1.337 \times {25.11 \times \sqrt{\frac{1}{7} +\frac{1}{11} } }$ , $(81.70-74.60)+1.337 \times {25.11 \times \sqrt{\frac{1}{7} +\frac{1}{11} } }$ ]

= [-9.132 , 23.332]

Therefore, 80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases ( $\mu_1-\mu_2$ ) is [-9.132 , 23.332].

4 0

3 years ago

Suppose that the functions and s are defined for all real numbers x as follows.

abruzzese [7]

(r*s) = (x+1)(3x+4) = 3x² + 3x + 4x + 4 = 3x² + 7x + 4

(r-s)(x) = x+1 - (3x+4) = x + 1 - 3x - 4 = -2x - 3

(r+s)(x) = x + 1 + 3x + 4 = 4x + 5

(r + s)(2) = 4(2) + 5 = 8 + 5 = 13

<h3>What is algebraic expression?</h3>

A number, a variable, or a mix of numbers, variables, and operation symbols make up an expression. Two expressions joined by an equal sign form an equation. Word illustration: The product of 8 and 3. Word illustration: The product of 8 and 3 is 11

to learn more about algebraic expression refer to:

brainly.com/question/4344214

#SPJ9

7 0

2 years ago