What are the solutions to the quadratic equation 3(x − 4)^2 = 75?

How to do a equation with 60 less than j. Would label it as 60--j or j--60 ??

slavikrds [6]

J -60
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3 0

2 years ago

John and his friend ordered lunch at a local sandwich shop. They each ordered a soft drink that costs $1.65. John orders a whole

ExtremeBDS [4]

Answer:

a. $12.54 = $ 1.65 + c/2 + c b. $ 7.26

Step-by-step explanation:

a. What equation can you use to find the cost c of a sandwich

Let c be the cost of a whole ham sandwich. Now, the total bill = cost of drinks + cost of half a ham sandwich + cost of a whole ham sandwich.

total bill = $ 12.54, cost of drinks = $ 1.65, cost of half a ham sandwich = c/2 and cost of a whole ham sandwich = c. So,

$12.54 = $ 1.65 + c/2 + c

b. What is the value of c

We then solve for c in the equation $12.54 = $ 1.65 + c/2 + c

collecting like terms, we have

$12.54 - $ 1.65 = c/2 + c

$ 10.89 = 3c/2

multiplying through by 2, we have

$ 10.89 × 2 = 3c

$ 21.78 = 3c

dividing through by 3, we have

c = $ 21.78/3

= $ 7.26

4 0

2 years ago

What is the answer to 2 - 6h when h = - 3

Oxana [17]

h=-3

Substitute h in 2-6h

2-6h=

2-6*(-3)=2+18=20

Answer: 20

6 0

3 years ago

Plz help me with this

seropon [69]

HEY mate here is your answer

I think the correct option is equivalent...

I am from India..

hope it helps you.

be brainly

8 0

3 years ago

Read 2 more answers

The attention span of children (ages 3 to 5) is claimed to be Normally distributed with a mean of 15 minutes and a standard devi

Agata [3.3K]

Answer:

If the observed sample mean is greater than 17.07 minutes, then we would reject the null hypothesis.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 15 minutes

Sample size, n = 10

Alpha, α = 0.05

Population standard deviation, σ = 4 minutes

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 15\text{ minutes}\\H_A: \mu > 15\text{ minutes}$

Since the population standard deviation is given, we use one-tailed z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Now, $z_{critical} \text{ at 0.05 level of significance for one tail} = 1.64$

Thus, we would reject the null hypothesis if the z-statistic is greater than this critical value.

Thus, we can write:

$z_{stat} = \displaystyle\frac{\bar{x} - 15}{\frac{4}{\sqrt{10}} } > 1.64\\\\\bar{x} - 15>1.64\times \frac{4}{\sqrt{10}}\\\bar{x} - 15>2.07\\\bar{x} > 15+2.07\\\bar{x} > 17.07$

Thus, the decision rule would be if the observed sample mean is greater than 17.07 minutes, then we would reject the null hypothesis.

6 0

3 years ago