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3 years ago

Minus then changes from 115.5 inches tall to 23.1 inches tall. What percent of change is that?

Mathematics

1 answer:

strojnjashka [21]3 years ago

4 0

Answer:

80%

Step-by-step explanation:

23.1/115.5=0.2

1-0.2=0.8=80%

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What is the sum of the geometric series 2^0 + 2^1 + 2^2 + 2^3 + 2^3 + 2^4 + … + 2^9?

GREYUIT [131]

Sum is
$S_{n}=\frac{a_{1}(1-r^{n})}{1-r}$

r=common ratio
a1=first term
it looks like 2^0=1 is the first term aka a1
it goes to the 9th term (2^9)

sub
$S_{9}=\frac{1(1-(2)^{9})}{1-2}$
$S_{9}=\frac{1-512}{-1}$
$S_{9}=\frac{-511}{-1}$
$S_{9}=511$

4 0

2 years ago

Read 2 more answers

Choose the correct answer 1 1/12 - 7/10 greater than half or less than half

CaHeK987 [17]

Answer:

less than half

Step-by-step explanation:

13/12 - 7/10 = 23/60

5 0

3 years ago

During the 7th examination of the Offspring cohort in the Framingham Heart Study, there were 1219 participants being treated for

AlexFokin [52]

Answer:

95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

Step-by-step explanation:

We are given that during the 7th examination of the Offspring cohort in the Framing ham Heart Study, there were 1219 participants being treated for hypertension and 2,313 who were not on treatment.

The sample proportion is : $\hat p$ = x/n = 1219/3532 = 0.345

Firstly, the pivotal quantity for 95% confidence interval for the proportion of the population is given by;

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

where, $\hat p$ = sample proportion = 0.345

n = sample of participants = 3532

p = population proportion

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of

significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

<u>95% confidence interval for p</u>= [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.345-1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } }$ , $0.345+1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } }$ ]