9,000 is 10 times greater than what number

A store sells Brazilian coffee for $10 per lb. and Columbian coffee for $14 per lb. If the store decides to make a 150-lb. blend

Artemon [7]

Answer:

The store should use 112.5 pounds of Brazilian coffee and 37.5 pounds of Colombian cofee.

Step-by-step explanation:

Let "b" be the amount of Brazilian coffee, in pounds, required for the blend and "c" the amount of Colombian coffee required, in pounds.

Since there are two unknown variables a two-equation system is needed to solve the problem, we can set up one equation for weight and another for price as follows:

$b+c=150\\10*b+14*c=11*150$

Solve for "c" by multiplying the first equation by -10 and adding it to the second one:

$b+c=150\\10b+14c -10b-10c=11*150 -(10*150)\\4c=150\\c = 37.5$

Now, solve for b by replacing the value obtained into the first equation

$b+c=150\\b= 150 - 37.5\\b=112.5$

The store should use 112.5 pounds of Brazilian coffee and 37.5 pounds of Colombian cofee.

6 0

3 years ago

Write the definition of a right angle in BICONDITIONAL form.*

VARVARA [1.3K]

Answer:

A biconditional statement is a combination of a conditional statement and its converse written in the if and only if form. Two line segments are congruent if and only if they are of equal length. ... A biconditional is true if and only if both the conditionals are true. there??

Step-by-step explanation:

5 0

3 years ago

What is the domain of f(x)=(2/3)^x

dlinn [17]

The domain of any exponential function is "all real numbers".

7 0

3 years ago

A web-based company has a goal of processing 95 percent of its orders on the same day they are received. If 485 out of the next

HACTEHA [7]

Answer:

The null hypothesis was rejected.

Conclusion: The proportions of orders completed on the day of receiving is more than 95%.

Step-by-step explanation:

The hypothesis can be defined as:

H₀: The proportions of orders completed on the day of receiving is not more than 95%, i.e. p ≤ 0.95.

Hₐ: The proportions of orders completed on the day of receiving is more than 95%, i.e. p > 0.95.

The significance level of the test is α = 0.025.

The sample size is, n = 500.

As the sample size is large, the sampling distribution of sample proportion can be approximated by the Normal distribution.

The mean and standard deviation of this distribution are:

$\mu=p\\\sigma=\sqrt{\frac{p(1-p)}{n}}$

The test statistic is:

$z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}} }$

The sample proportion is:

$\hat p=\frac{X}{n}=\frac{485}{500}=0.97$

Compute the test statistic as follows:

$z=\frac{0.97-0.95}{\sqrt{\frac{0.95(1-0.95)}{500}} }=2.05$

The decision rule is:

If the p-value is less than the significance level α then the null hypothesis is rejected.

Compute the p-value as follows:

$p-value=P(Z>2.05)\\=1-P(Z$

*Use a z-table.

The p-value = 0.0202 < α = 0.025.

The null hypothesis will be rejected.

Conclusion:

As the null hypothesis was rejected at 2.5% level of significance, it can be concluded that the proportions of orders completed on the day of receiving is more than 95%.

3 0

3 years ago

Read 2 more answers

The screening process for detecting a rare disease is not perfect. Researchers have developed a blood test that is considered fa

lesya692 [45]

Answer:

Type I: 1.9%, Type II: 1.6%

Step-by-step explanation:

given null hypothesis

H0=the individual has not taken steroids.

type 1 error-falsely rejecting the null hypothesis

⇒ actually the null hypothesis is true⇒the individual has not taken steroids.

but we rejected it ⇒our prediction is the individual has taken steroids.

typr II error- not rejecting null hypothesis when it has to be rejected

⇒actually null hypothesis is false ⇒the individual has taken steroids.

but we didnt reject⇒the individual has not taken steroids.

let us denote

the individual has taken steroids by 1

the individual has not taken steroids.by 0

predicted

1 0

actual 1 98.4% 1.6%

0 1.9% 98.1%

so for type 1 error

actual-0

predicted-1

therefore from above table we can see that probability of Type I error is 1.9%=0.019

so for type II error

actual-1

predicted-0

therefore from above table we can see that probability of Type I error is 1.6%=0.016

5 0

3 years ago