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guapka [62]
3 years ago
14

Please someone if you know how to do this I need help 10. And 11 please.

Mathematics
2 answers:
Effectus [21]3 years ago
5 0

Answer:

10.

Bag A is LIGHTER than Bag B

Bag B is LIGHTER than Bag C

So, Bag A is LIGHTER than Bag C

11.

pear is the heaviest, and the strawberry is the lightest

Step-by-step explanation:

This is really simple! Looking at the weight scale, the heavier object always goes lower than the lighter object. Knowing this, we can figure out which is heavier and which is lighter.

10.

Bag A is LIGHTER than Bag B

Bag B is LIGHTER than Bag C

Looking at these we can infer that bag c is heavier than both.

So, Bag A is LIGHTER than Bag C

11. The pear falls lower than the apple, meaning it is heavier. The apple falls lower, meaning is it heavier than the apple. We can infer that the pear is the heaviest, and the strawberry is the lightest

Hope this helps!!

-Ketifa

artcher [175]3 years ago
3 0

Answer:

10.

bag a is lighter than bag b

bag b is lighter than bag c

so bag a is lighter that bag c

Step-by-step explanation:

... sorry i dont know #11, but hope this helps

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4 years ago
A university financial aid office polled a random sample of 670 male undergraduate students and 617 female undergraduate student
ahrayia [7]

Answer:

The 90% confidence interval for the difference between the proportions of male and female students who were employed during the summer is (0.01, 0.1012).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Male undergraduates:

670, 388 were employed. So

p_M = \frac{388}{670} = 0.5791

s_M = \sqrt{\frac{0.5791*0.4209}{670}} = 0.0191

Female undergraduates:

Of 617, 323 were employed. So

p_F = \frac{323}{617} = 0.5235

s_F = \sqrt{\frac{0.5235*0.4765}{617}} = 0.0201

Distribution of the difference:

p = p_M - p_F = 0.5791 - 0.5235 = 0.0556

s = sqrt{s_M^2+s_F^2} = \sqrt{0.0201^2 + 0.0191^2} = 0.0277

Confidence interval:

The confidence interval is given by:

p \pm zs

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

90% confidence level

So \alpha = 0.1, z is the value of Z that has a p-value of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

The lower bound of the interval is:

p - zs = 0.0556 - 1.645*0.0277 = 0.01

The upper bound of the interval is:

p + zs = 0.0556 + 1.645*0.0277 = 0.1012

The 90% confidence interval for the difference between the proportions of male and female students who were employed during the summer is (0.01, 0.1012).

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variation\\
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