All categories

4 years ago

Write two ways you could use to find 5+8

Mathematics

2 answers:

Gre4nikov [31]4 years ago

7 0

Well, you can do, 4+9, and 5+9, or, instead of 5+9, maybeeeee, 6+7!

Tatiana [17]4 years ago

4 0

5+9 thats one way....buts theres alot of ways to get 13
Hope it helps

<3

You might be interested in

How many two-digit positive integer less than 25 have the property that they are divisible by each of their digits? List themZ

Ierofanga [76]

This is easy, since there aren't many positive two digit integers under 25.
Just go through and check each one!

10 - not divisible by 0.
11 - divisible
12 - divisible
13 - not divisible by 3.
14 - not divisible by 4.
15 - divisible
16 - not divisible by 6.
17 - not divisible by 7.
18 - not divisible by 8.
19 - not divisible by 9.
20 - not divisible by 0.
21 - not divisible by 2.
22 - divisible
23 - not divisible by 2 or 3.
24 - divisible

To sum it up, the numbers 11, 12, 15, 22, and 24 are divisible by each of their digits.

Hope this helps (and your fortune is "Good things will come your way if you keep eating pizza.")!

8 0

4 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.

Let $\mu_1$ = true mean number of sales at location A.

$\mu_2$ = true mean number of sales at location B

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 Two-sample t test statistics 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, test statistics = $\frac{(39-49)-(0)}{6.64 \times \sqrt{\frac{1}{18}+\frac{1}{13} } }$ ~ $t_2_9$

= -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.

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 we reject our null hypothesis.

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

All of the following proportions are equivalent except a/b=c/d b/a=d/c a/d=c/b a/c=b/d

valentina_108 [34]

I believe the correct answer from the choices listed above is the third option. All of the following proportions are equivalent except a/d=c/b.

a/b=c/d

b/a=d/c => a/b = c/d

a/d=c/b not equal to a/b = c/d

a/c=b/d => a/b = c/d

7 0

3 years ago

If the sides of a cube are 2 meters long, then its volume is A) 8 m B) 8 m2 C) 8 m3 D) 8 m4 Eliminate

grigory [225]