-25-1/2= answer choices a) -5 b) -1/5 c) no real number

Mathematics

2 answers:

suter [353]3 years ago

6 0

Answer:

C) no real number

Step-by-step explanation:

MakcuM [25]3 years ago

3 0

Answer:

The Correct answer is -1/5

Step-by-step explanation:

the equation above, is -1/5 when simplified. Other answers could be -0.2, -5-1

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1/8% is what as a decimal

vlabodo [156]

Answer:

$\large\boxed{\dfrac{1}{8}\%=\dfrac{1}{800}}$

Step-by-step explanation:

$p\%=\dfrac{p}{100}\\\\\dfrac{1}{8}\%=\dfrac{\frac{1}{8}}{100}=\dfrac{1}{8}\cdot\dfrac{1}{100}=\dfrac{1}{800}\\\\\text{other method}\\\\\text{Convert the fraction to the decimal:}\\\\\dfrac{1}{8}=0.125\to\dfrac{1}{8}\%=0.125\%\\\\0.125\%=\dfrac{0.125}{100}=\dfrac{0.125\cdot1000}{100\cdot1000}=\dfrac{125}{100000}=\dfrac{125:125}{100000:125}=\dfrac{1}{800}$

3 0

3 years ago

Find the confidence interval with a 90% level of confidence. In October 2010, Gallup Poll asked 510 randomly sampled adults the

ratelena [41]

Answer:

90% confidence interval for the percent of the population who consider the death penalty to be "fair" is [54.4% , 61.6%].

Step-by-step explanation:

We are given that in October 2010, Gallup Poll asked 510 randomly sampled adults the question, "Generally speaking, do you believe the death penalty is applied fairly or unfairly in this country today?"

Of these, 58% answered "fairly", 36% said "unfairly" and 7% said they didn't know.

Firstly, the Pivotal quantity for 90% confidence interval for the population proportion 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 who answered death penalty "fair" = 58%

n = sample of adults = 510

p = population percentage

Here for constructing 90% confidence interval we have used One-sample z test for proportions.

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

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

90% confidence interval for p = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.58-1.645 \times {\sqrt{\frac{0.58(1-0.58)}{510} } }$ , $0.58+1.645 \times {\sqrt{\frac{0.58(1-0.58)}{510} } }$ ]