All categories

3 years ago

Total cost for cookie and lemonade is 1.80. If lemonade costs 1.00 more than the cookie, how much was the lemonade?

Mathematics

1 answer:

Ray Of Light [21]3 years ago

5 0

Answer:

The lemonade costs 1.4.

Step-by-step explanation:

This question is solved using a system of equations.

I am going to say that:

x is the cost of a cookie.

y is the cost of a lemonade.

Total cost for cookie and lemonade is 1.80.

This means that $x + y = 1.8$ .

As we want y, we have that $x = 1.8 - y$

Lemonade costs 1.00 more than the cookie?

This means that:

$y = x + 1$

$y = 1.8 - y + 1$

$2y = 2.8$

$y = \frac{2.8}{2}$

$y = 1.4$

The lemonade costs 1.4.

You might be interested in

Please help me !!!!!!!!

LuckyWell [14K]

51.3 I think you just have to subtract 68.5 to 17.2

7 0

2 years ago

Which expression is equivalent to -32 to 3/5 power

levacccp [35]

Answer:

Step-by-step explanation:

We are given the following expression and we are to find the simplest form of this expression:

$- 3 2 ^ { \frac { 3 } { 5 } }$

First of all, we will factor the coefficient:

$3 2 = 2 ^ 5$

Rewriting it as:

$- ( 2 ^ 5 ) ^ { \frac { 3 } { 5 } }$

Applying the exponent rule $(a^b)^c = a^{bc}$ to get:

$-2^{5.\frac{3}{5}}=-2^3$

$-2^3=-8$

6 0

3 years ago

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 48,564 miles, with a standard

DerKrebs [107]

Answer:

0.0091 = 0.91% probability that the sample mean would be less than 48,101 miles in a sample of 281 tires if the manager is correct

Step-by-step explanation:

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

Normal probability distribution:

Problems of normally distributed samples can be 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 $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$