When calculating how much you saved what do you have to do

Mathematics

1 answer:

Vladimir [108]3 years ago

4 0

Answer:

$\frac{original\ price\ -\ discounted\ price}{original\ price}*100\%$

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Can you help me i dont understand it?

hram777 [196]

208 in^2

find the area of each of the parts (4 triangles and 1 square)

all of the triangles have the same area since this is a square pyramid
area of triangle = (base X height)/2
so it would be (8 X 9)/2 = 36

for the square, the area is length X width
so it would be 8 X 8 = 64

add all the areas
36+36+36+36+64 = 208

8 0

3 years ago

Read 2 more answers

To create an entry code you must first choose 3 letters and then 2 single digit numbers how many different entry codes can u cre

horrorfan [7]

3!(2!)=
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12

6 0

3 years ago

We wish to estimate what percent of adult residents in a certain county are parents. Out of 300 adult residents sampled, 144 had

ser-zykov [4K]

Answer:

90% confidence interval for the proportion p of adult residents who are parents in this county is [0.433 , 0.527].

Step-by-step explanation:

We are given that we wish to estimate what percent of adult residents in a certain county are parents. Out of 300 adult residents sampled, 144 had kids.

Firstly, the pivotal quantity for 90% confidence interval for the proportion p of adult residents who are parents in this county is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = proportion of adults residents who are parents in this county in a sample of 300 adults = $\frac{144}{300}$ = 48%

n = sample of adults residents = 300

p = population proportion of adults

<em>Here for constructing 90% confidence interval we have used One-sample z proportion statistics.</em>

So, 90% confidence interval for the population proportion, p is ;

P(-1.6449 < N(0,1) < 1.6449) = 0.90 {As the critical value of z at 5%

significance level are -1.6449 & 1.6449}

P(-1.6449 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.6449) = 0.90

P( $-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

<u>90% confidence interval for p </u>= [ $\hat p-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48-1.6449 \times {\sqrt{\frac{0.48(1-0.48)}{300} } }$ , $0.48+1.6449 \times {\sqrt{\frac{0.48(1-0.48)}{300} } }$ ]