Ask question

Ask question

All categories

3 years ago

7

Which quadrilateral has diagonals that always bisect its angles and also bisect each other? 1) rhombus.2)rectangle.3)parallelogr

am.4)isosceles trapezoid.

2 answers:

Juliette [100K]3 years ago

6 0

The answer is 2. rectangles.
Hope this helps!

Art [367]3 years ago

5 0

I want to say parallelogram

You might be interested in

HELP ME ASAP!!! PLEASE FAST!

ryzh [129]

Blank #1 is 0.67 and Blank #2 is 13.3. I am not sure about Blank #3. Here is a tip: Mean absolute deviation is the average of the absolute deviations. Tell me if I am right ok?

7 0

3 years ago

Jessica and Emily each improved their yards by planting daylilies and ivy. They bought their supplies from the same store. Jessi

Darya [45]

Answer:The cost of one daylily is 6, and the cost of one pot of ivy is 8.

Step-by-step explanation:

7 0

3 years ago

When multiplying fractions do you multiple it like this: 5/6 × 7/8 35/48 and if you are multiplying fractions with the same deno

Mnenie [13.5K]

Step-by-step explanation:

5/8 × 6/8= 30/8 => this is the wrong one

it should be 5/8 × 6/8= 30/64.

9/10×3/10= 27/100 => this is the true one

6 0

3 years ago

Location is known to affect the number, of a particular item, sold by an automobile dealer. Two different locations, A and B, ar

yKpoI14uk [10]

Answer:

We conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

Step-by-step explanation:

We are given that Location A was observed for 18 days and location B was observed for 13 days.

On average, location A sold 39 of these items with a sample standard deviation of 8 and location B sold 49 of these items with a sample standard deviation of 4.

<em>Let </em> $\mu_1$ <em> = true mean number of sales at location A.</em>

<em /> $\mu_2$ = <em>true mean number of sales at location B</em>

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2\geq0$ or $\mu_1 \geq \mu_2$ {means that the true mean number of sales at location A is greater than or equal to the true mean number of sales at location B}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2$ or $\mu_1< \mu_2$ {means that the true mean number of sales at location A is fewer than the true mean number of sales at location B}

The test statistics that would be used here <u>Two-sample t test statistics</u> as we don't know about the population standard deviations;

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 average of items sold at location A = 39

$\bar X_2$ = sample average of items sold at location B = 49

$s_1$ = sample standard deviation of items sold at location A = 8

$s_2$ = sample standard deviation of items sold at location B = 4

$n_1$ = sample of days location A was observed = 18

$n_2$ = sample of days location B was observed = 13

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(18-1)\times 8^{2}+(13-1)\times 4^{2} }{18+13-2} }$ = 6.64

So, <u><em>test statistics</em></u> = $\frac{(39-49)-(0)}{6.64 \times \sqrt{\frac{1}{18}+\frac{1}{13} } }$ ~ $t_2_9$

= -4.14

The value of t test statistics is -4.14.

Now, at 0.01 significance level the t table gives critical value of -2.462 at 29 degree of freedom for left-tailed test.

<em>Since our test statistics is less than the critical values of t as -2.462 > -4.14, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>

Therefore, we conclude that the true mean number of sales at location A is fewer than the true mean number of sales at location B.

3 0

3 years ago

What is the solution of the equation? Enter your answer, as a fraction, in the box

Effectus [21]

X= 2/3 hope that helps

5 0

3 years ago