A recipe calls for 3 3/4 cups of flour how much flour is needed to make 1/2 the recipe

The cost of 60 shares of fly by night airlines is $67.50. How many shares can you buy with $675.00?

crimeas [40]

Answer:

10 shares.

Step-by-step explanation:

So first you need to right yourself an equation to help you solve this

so it cost $67.50 with 60 shares.

so let x be the shares

67.50x=675.00

divide 67.50 by 675.00 and then you should get your answer of 10 shares.

the answer therefore is 10 shares.

7 0

2 years ago

When lit, a candle loses 8% of the unlit candle every hour. Janet lit a candle with an

sweet [91]

Answer: 2.5 hours

Step-by-step explanation:

8 0

2 years ago

The manager of a warehouse would like to know how many errors are made when a product's serial number is read by a bar-code read

Bond [772]

Answer:

$(a)\ \bar x =20.5$

$(b)\ \sigma = 14.54$

Step-by-step explanation:

Given

$x: 36, 14, 21, 39, 11, 2$

Solving (a): The mean of the 6 samples.

This is calculated as:

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x =\frac{36+ 14+ 21+ 39+ 11+ 2}{6}$

$\bar x =\frac{123}{6}$

$\bar x =20.5$

Solving (b): The sample standard deviation.

This is calculated using:

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}$

So, we have:

$\sigma = \sqrt{\frac{(36 - 20.5)^2 + (14 - 20.5)^2 + (21 - 20.5)^2 + (39 - 20.5)^2 + (11 - 20.5)^2 + (2 - 20.5)^2}{6-1}}$

$\sigma = \sqrt{\frac{1057.5}{5}}$

$\sigma = \sqrt{211.5}$

$\sigma = 14.54$

7 0

3 years ago

Tony’s investments lost $12,545.50 in value over the past 2.5 months. On average, how much did his investments change each month

zvonat [6]

Answer:

Step-by-step explanation: 25÷ 12545.5 = 5,018.20

8 0

2 years ago

A physicist examines 25 water samples for nitrate concentration. The mean nitrate concentration for the sample data is 0.165 cc/

xz_007 [3.2K]

Answer:

The 80% confidence interval for the the population mean nitrate concentration is (0.144, 0.186).

Critical value t=1.318

Step-by-step explanation:

We have to calculate a 80% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=0.165.

The sample size is N=25.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{0.078}{\sqrt{25}}=\dfrac{0.078}{5}=0.016$

The degrees of freedom for this sample size are:

$df=n-1=25-1=24$

The t-value for a 80% confidence interval and 24 degrees of freedom is t=1.318.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=1.318 \cdot 0.016=0.021$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 0.165-0.021=0.144\\\\UL=M+t \cdot s_M = 0.165+0.021=0.186$

The 80% confidence interval for the population mean nitrate concentration is (0.144, 0.186).

6 0

3 years ago