Ask question

Ask question

All categories

2 years ago

9

The Spanish club sold hot pretzels as a fundraiser. The pretzels normally sold for 2.00 but near the end of the sale the price w

as reduced by 25%. What was the new price for a hot pretzel

1 answer:

Degger [83]2 years ago

6 0

25% of 2$ is 50cents, so the new price was 1.50$

You might be interested in

Factor the polynomial by its greatest common monomial factor.

Gwar [14]

Answer: 3b^4(b+5-6b^3)

Step-by-step explanation:

6 0

2 years ago

Given that 1 x2 dx 0 = 1 3 , use this fact and the properties of integrals to evaluate 1 (4 − 6x2) dx. 0

Debora [2.8K]

So, the definite integral $\int\limits^1_0 {(4 - 6x^{2} )} \, dx= - 74$

Given that

$\int\limits^1_0 {x^{2} } \, dx = 13$

We find

$\int\limits^1_0 {(4 - 6x^{2} )} \, dx$

<h3>Definite integrals </h3>

Definite integrals are integral values that are obtained by integrating a function between two values.

So, $Integral \int\limits^1_0 {(4 - 6x^{2} )} \, dx$

So, $\int\limits^1_0 {(4 - 6x^{2} )} \, dx = \int\limits^1_0 {4} \, dx - \int\limits^1_0 {6x^{2} } \, dx \\= 4[x]^{1}_{0} - \int\limits^1_0 {6x^{2} } \, dx \\= 4[x]^{1}_{0} - 6\int\limits^1_0 {x^{2} } \, dx \\= 4[1 - 0] - 6\int\limits^1_0 {x^{2} } \, dx\\= 4[1] - 6\int\limits^1_0 {x^{2} } \, dx\\= 4 - 6\int\limits^1_0 {x^{2} } \, dx$

Since

$\int\limits^1_0 {x^{2} } \, dx = 13$ ,

Substituting this into the equation the equation, we have

$\int\limits^1_0 {(4 - 6x^{2} )} \, dx = 4 - 6\int\limits^1_0 {x^{2} } \, dx\\= 4 - 6 X 13 \\= 4 - 78\\= -74$

So, $\int\limits^1_0 {(4 - 6x^{2} )} \, dx= - 74$

Learn more about definite integrals here:

brainly.com/question/17074932

4 0

2 years ago

What is the domain of g(x) = sqrt x +1

Shkiper50 [21]

There can be no negative in a sqrt
so x+1 must be equal or greater then 0
x+1 $\geq$ 0
x $\geq$ -1
so all answers must be greater then or equal to -1
-1,0,1,2,3,4,5,6,7,8
(-1,infinity) is the domain

7 0

2 years ago

A sample of 1500 computer chips revealed that 32% of the chips do not fail in the first 1000 hours of their use. The company's p

Grace [21]

Answer:

Yes, we have sufficient evidence at the 0.02 level to support the company's claim.

Step-by-step explanation:

We are given that a sample of 1500 computer chips revealed that 32% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed that above 29% do not fail in the first 1000 hours of their use.

Let Null Hypothesis, $H_0$ : p $\leq$ 0.29 {means that less than or equal to 29% do not fail in the first 1000 hours of their use}

Alternate Hypothesis, $H_1$ : p > 0.29 {means that more than 29% do not fail in the first 1000 hours of their use}

The test statics that will be used here is One-sample proportions test;

T.S. = $\frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = proportion of chips that do not fail in the first 1000 hours of their use = 32%

n = sample of chips = 1500

So, <u>test statistics</u> = $\frac{0.32-0.29}{\sqrt{\frac{0.32(1-0.32)}{1500} } }$

= 2.491

<em>Now, at 0.02 level of significance the z table gives critical value of 2.054. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject null hypothesis as it fall in the rejection region.</em>

Therefore, we conclude that more than 29% do not fail in the first 1000 hours of their use which means we have sufficient evidence at the 0.02 level to support the company's claim.

7 0

3 years ago

Three fair coins are tossed. If all land "heads," the player wins $10, and if exactly two land heads, the player wins $5. If it

hodyreva [135]

Answer:

<u>After four games, a player can lose up to $ 16 to win up to $ 26. These are the probabilities for every game:</u>

<u>1/8 or 12.5% of landing three "heads"</u>

<u>3/8 or 37.5% of landing two "heads"</u>

<u>4/8 or 50% of landing no or only one "head".</u>

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

If three coins land "heads" the player wins $ 10

If two coins land "heads" the player wins $ 5

Cost of playing = $ 4

2. What is the player's expected outcome after four games?

Probability of two coins out of three lands "heads" = 3/8

Probability of three coins out of three lands "heads" = 1/8

Now, let's calculate the player's expected outcome, as follows:

Four games:

Cost = 4 * 4 = $ 16

Worst-case scenario: No wins

Best-case scenario: 4 out of 4 of $ 10 win

Worst-case scenario profit or loss = 0 - 16 = Loss of $ 16

Best-case scenario profit or loss = 40 - 16 = Profit of $ 24

After four games, a player can lose up to $ 16 to win up to $ 26. These are the probabilities for every game:

1/8 or 12.5% of landing three "heads"

3/8 or 37.5% of landing two "heads"

4/8 or 50% of landing no or only one "head".

8 0

2 years ago