Can someone help me as soon as possible!

How many combinations are possible? Assume the items are distinct.<br> 10 items chosen 10 at a time

elena-14-01-66 [18.8K]

About 100 combinations are possible because u have to do 10x10

3 0

3 years ago

Which shows one way the equation can be represented in words?

Anuta_ua [19.1K]

Answer:

C

Step-by-step explanation:

8 0

3 years ago

On a map, distance in km and distance in cm varies directly, and 25 km are represented by 2cm. If two cities

ikadub [295]

Answer:

87.5km

Step-by-step explanation:

We can create an equation to represent what 1 cm on the map is.

$2cm = 25km\\1cm = 12.5km\\$

Now that we know what 1 cm is, we can multiply both sides by 7 to get the actual distance in km.

$1cm * 7 = 12.5km * 7\\7cm = 87.5km$

5 0

3 years ago

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Could someone help me with this (picture) please explain if you could I don't really understand.

gregori [183]

Answer:

10t + D = 40

Step-by-step explanation:

y = mx + b is the equation of a line in slope intercept form. We'll be using this equation since the slope, or distance traveled, is steady (linear). In this case, the slope is 10 since Sarah's pace is 10 km/hr. m is the slope, so m = 10.

y = 10x + b. However, she needs to run 40 km. So we're going to substitute 40 in for y.

40 = 10x + b. Now, we need to use t instead of x since that's what's been asked of us. (The slope represents the distance traveled over time, so t fits.)

10t + b = 40

b needs to become D, the distance in kilometers.

So now we have 10t + D = 40

7 0

3 years ago

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A study tested whether significant social activities outside the house in young children affected their probability of later dev

Alona [7]

Answer:

Step-by-step explanation:

From the question we are told that

The first sample size is $n_1 = 1000$

The second sample size is $n_2 = 6000$

The number that had significant outside activity in the sample with ALL is $k_1 = 700$

The number that had significant outside activity in the sample without ALL is $k_2 = 5000$

Considering question a

The percentage of children with ALL have significant social activity outside the home when younger is mathematically represented as

$\^ p_1 = \frac{700}{1000} * 100$

=> $\^ p_ 1 = 0.7 = 70\%$

Considering question b

The percentage of children without ALL have significant social activity outside the home when younger is mathematically represented as

$\^ p_2 = \frac{5000}{6000}$

=> $\^ p_ 2 = 0.83$

Generally the sample odds ratio for significant social activity outside the home when younger, comparing the groups with and without ALL is mathematically represented as

$r = \frac{\* p _1}{ \^ p_2 }$

=> $r = \frac{0.7}{ 0.83 }$

=> $r = 0.141$

Considering question c

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the lower limit of the 95% confidence interval for the population odds ratio for significant social activity outside the home when younger, comparing the groups with and without ALL is mathematically represented as

$a = e^{ln ( r ) - Z_{\frac{\alpha }{2}} \sqrt{ [ \frac{1}{ k_1 } ] + [ \frac{1}{ c_1 } ] + [\frac{1}{k_2} ] + [\frac{1}{ c_2 } ] } }$

Here $c_1 \ and \ c_2$ are the non-significant values i.e people that did not play outside when they were young in both samples

The values are

$c_1 = 1000 - 700 = 300$

and $c_2 = 6000 - 5000$

=> $c_2 = 1000$

=> $a = e^{ln ( 0.141 ) - 1.96 \sqrt{ [ \frac{1}{ 700 } ] + [ \frac{1}{ 1000} ] + [\frac{1}{5000} ] + [\frac{1}{ 300 } ] } }$

=> $a = 0.1212$

Generally the upper limit of the 95% confidence interval for the population odds ratio for significant social activity outside the home when younger, comparing the groups with and without ALL is mathematically represented as

$b = e^{ln ( 0.141 ) + 1.96 \sqrt{ [ \frac{1}{ 700 } ] + [ \frac{1}{ 1000} ] + [\frac{1}{5000} ] + [\frac{1}{ 300 } ] } }$

$b = 0.1640$

Generally the 95% confidence interval for the population odds ratio for significant social activity outside the home when younger, comparing the groups with and without ALL is

$95\% CI = [ 0.1212 , 0.1640 ]$

Generally looking and the confidence interval obtained we see that it is less that 1 hence this means that there is a greater odd of developing ALL in groups with insignificant social activity compared to groups with significant social activity

3 0

3 years ago

Can someone help me as soon as possible!​

Can someone help me as soon as possible!