Find the size of the final unknown exterior angle in a polygon whose

Compare the values of the underlined digits. 503497 and 26475. The underlined is 7

shtirl [24]

The 7 is 10 times bigger in 26475 than the 7 in 503497. why? because the 7 in 26475 is in the 10's place and the 7 in 503497 is in the ones place.

If you need clarification on the 10 times bigger part...consider this whats 1×10? 10
. If the 7 was in the 100 place and the other 7 was in the 10 place then it would be 10×10=100 and so on

3 0

3 years ago

Jon recently drove to visit his parents who live 270 270 miles away. On his way there his average speed was 24 24 miles per hour

Ber [7]

Answer:

He drove there at 60 mph, and he drove back at 36 mph.

Step-by-step explanation:

one way distance = d = 270 miles

average speed on way back = s

average speed on the way there = s + 24

time driving there = t

time driving back = 12 - t

average speed = distance/time

distance = speed * time

going there:

270 = (s + 24)t

270 = st + 24t

going back

270 = s(12 - t)

270 = 12s - st

We have a system of equations:

270 = st + 24t

270 = 12s - st

Solve the first equation for t.

t(s + 24) = 270

t = 270/(s + 24)

Substitute in the second equation.

270 = 12s - s[270/(s + 24)]

270 = 12s - 270s/(s + 24)

Multiply both sides by s + 24.

270s + 6480 = 12s^2 + 288s - 270s

12s^2 - 252s - 6480 = 0

Divide both sides by 12.

s^2 - 21s - 540 = 0

(s - 36)(s + 15) = 0

s = 36 or s = -15

The average speed cannot be negative, so we discard the solution s = -15.

s = 36

s + 24 = 60

Answer: He drove there at 60 mph, and he drove back at 36 mph.

3 0

2 years ago

Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 117 mg/dl and

TiliK225 [7]

Answer:

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

Step-by-step explanation:

To solve this problem, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 117, \sigma = 10.6, n = 6, s = \frac{10.6}{\sqrt{6}} = 4.33$

What is the level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L ?

This is the value of X when Z has a pvalue of 1-0.01 = 0.99. So X when Z = 2.33.

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$2.33 = \frac{X - 117}{4.33}$

$X - 117 = 2.33*4.33$

$X = 127.1$

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

6 0

3 years ago

The area of a rectangular garden is given by the trinomial x2 + x – 42. What are the possible dimensions of the rectangle? Use f

Murrr4er [49]

Answer:

Step-by-step explanation:

Hello, please consider the following.

$x^2+x-42\\\\\text{The sum of the zeroes is -1=-7+6 and the product is -42=-7*6.}\\\\\text{So, we can factorise.}\\\\x^2+x-42\\\\=x^2+7x-6x-42\\\\=x(x+7)-6(x+7)\\\\=(x+7)(x-6)$

So, the possible dimensions of the rectangle are (x+7) and (x-6).

Thank you.

3 0

3 years ago

Read 2 more answers

A consumer group is interested in estimating the proportion of packages of ground beef sold at a particular store that have an a

storchak [24]

Answer: 267.

Step-by-step explanation:

When there is no prior information for the population proportion, then the formula we use to find the sample size to estimate the confidence interval :

$n=0.25(\dfrac{z^*}{E})^2$ , where z* = Critical z-value and E + amrgin of error.

Let p = proportion of packages of ground beef sold at a particular store that have an actual fat content exceeding the fat content stated on the label.

Since , we have no prior information about p. so we use above formula

with E = 0.06 and critical value for 95% confidence =z* =1.96 [By z-table ] , we get

$n=0.25(\dfrac{1.96}{0.06})^2=0.25(32.6666)^2\\\\=(0.25)(1067.111111)\\\\=266.777777778\approx267$

Hence, the required sample size is 267.

7 0

2 years ago