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2 years ago

There are 20 baseball cards and 10 football cards on display at a sports museum. Noah says that

Mathematics

1 answer:

erica [24]2 years ago

6 0

Answer:

They are both right becuase if you just take away 1 from 20 and 10 you get 2 and 1. For the 4:2 one take away 1 and mutiply by 2 equalling 4:2 but they are still the same thing.

Step-by-step explanation:

You might be interested in

The Girl Scouts collected $90 for charity while the Boy Scouts collected $60.

Ostrovityanka [42]

So first you’ll have to find the total or the sum of collected.

$90+$60= $150

Because you’re finding the percent of the total collected by girl scouts, divide the amount collected from the girl scouts by the total.

$90/$150= .6

Multiply that by 100 to get %60.

6 0

2 years ago

pls help no one will how do estimate the numbers im about to give you by rounding to the nearest integer 15.23 - 6.835 and 6.88

gayaneshka [121]

I think it's just 15, -7, 7, -8, and 80.
I am probably wrong though,
Is it all one long equation with the plus and minus signs?

8 0

3 years ago

39,012 in expanded form

Helen [10]

30,000 + 9,000 + 10 + 2

6 0

3 years ago

Find the slope of the line graphed below. İ -2 -1 3 1

Sonja [21]

Answer:

slope: 2/5

Step-by-step explanation:

5 0

3 years ago

Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r

ZanzabumX [31]

Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean weight = $\frac{\sum X}{n}$ = 21.88 ounces

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 0.201 ounces

n = sample of boxes = 12

$\mu$ = population mean weight

Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.796 < $t_1_1$ < 1.796) = 0.90 {As the critical value of t at 11 degrees of

freedom are -1.796 & 1.796 with P = 5%}

P(-1.796 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.796) = 0.90

P( $-1.796 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $21.88-1.796 \times {\frac{0.201}{\sqrt{12} } }$ , $21.88+1.796 \times {\frac{0.201}{\sqrt{12} } }$ ]

= [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

8 0

3 years ago