The school wishes to purchase 2500 pens and have discovered that 3 stores sell the same model pen.Which store has the best buy ?

Use the formula A=2πrh to find the area of the curved surface of each of the cylinders below. (Express your answers correct to 1

shusha [124]

Answer:

here,

A=2×22÷7×17/2×21

A=22×17×3

A=1122 sq.cm

3 0

3 years ago

The Royal Fruit Company produces two types of fruit drinks. The first type is 60% pure fruit juice, and the second type is 85% p

elena-s [515]

Make 2 equations from the question first
x is the number of pints for type 1
y is the number of pints for type 2

The equation
x + y = 120
60% x + 85% y = 65% (x + y)

Solve the equation
From the 2nd equation
0.6x + 0.85y = 0.65(x + y)
0.6x + 0.85y = 0.65x + 0.65y
0.85y - 0.65y = 0.65x - 0.6x
0.2y = 0.05x
y = 4x

From the 1st equation
x + y = 120
x + 4x = 120
5x = 120
x = 24

y = 4x
y = 96

The first type should be 24 pints, the second type should be 96 pints

5 0

3 years ago

A machine that fills beverage cans is supposed to put 12 ounces of beverage in each can. The standard deviation of the amount in

erica [24]

Answer:

$\chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54$

The degrees of freedom are:

$df=n-1=10-1=9$

Now we can calculate the p value using the alternative hypothesis:

$p_v =2*P(\chi^2 >13.54)=0.279$

Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces

Step-by-step explanation:

Assuming the following data:"12.14 12.05 12.27 11.89 12.06

12.14 12.05 12.38 11.92 12.14"

We can calculate the sample deviation with this formula:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$n=10$ represent the sample size

$\alpha=0.05$ represent the confidence level

$s^2 =0.0217$ represent the sample variance

$\sigma^2_0 =0.12^2= 0.0144$ represent the value to verify

Null and alternative hypothesis

We want to determine whether the standard deviation differs from 0.12 ounce, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 = 0.0144$

Alternative hypothesis: $\sigma^2 \neq 0.0144$

The statistic can be calculated like this;

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

$\chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54$

The degrees of freedom are:

$df=n-1=10-1=9$

Now we can calculate the p value using the alternative hypothesis:

$p_v =2*P(\chi^2 >13.54)=0.279$

Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces

7 0

3 years ago

Professor Vinculum investigated the migration season of the Bulbul bird from their natural wetlands to a warmer climate. He foun

coldgirl [10]

The population of the Bulbul birds at the start of the migration season is 1800.

<h3>Exponential function</h3>

An exponential function is given by:

y = abˣ

Let y, x are variables, a is the initial value and b is the multiplication factor.

Let P be the population of the bird t days since the start of the migration season.

$P=1350+400(1.25)^{-t}\\\\At\ the\ start\ of\ the\ season(t=0):\\\\P=1350+400(1.25)^{-0}\\\\P=1350+450=1800$

The population of the Bulbul birds at the start of the migration season is 1800.

Find out more on Exponential function at: brainly.com/question/12940982

8 0

2 years ago

Triangle S and Triangle L are scaled copies of one another. Type answers to each question in the boxes below. What is the scale

aev [14]

Answer:

Follows are the solution to this question:

Step-by-step explanation:

In the given-question, the missing data is defined in the attached file, please find it.

The answer is " $\bold{2, \ \frac{1}{2}, \ and \ \frac{2}{3}}$ ".

4 0

3 years ago