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1 year ago

Which statement describes a process to solve sqrt b+20 - sqrt b=5?

Mathematics

1 answer:

babunello [35]1 year ago

7 0

Answer:

Step-by-step explanation:

The equation is:

√b+20 - √b = 5

The first step is we will add √b to both sides:

√b+20 -√b +√b = 5 +√b

√b+20 = 5+√b

Now take square at both sides:

(√b+20)^2 = (5+√b)^2

b+20 = 25+10√b+b

Now combine the like terms:

b+20-25-b=10√b

-5 = 10√b

Divide both the terms by 10

-5/10 = 10√b/10

-1/2=√b

Take square at both sides:

(-1/2)^2 = (√b)^2

1/4 = b

So in this type of question we add radical terms to both sides and square both sides twice....

You might be interested in

In a USA TODAY/Gallup Poll, respondents favored Barack Obama over Mitt Romney in terms of likeability, 61% to 32% (Los Angeles T

dybincka [34]

Answer:

a) percentage of respondents that favored neither Obama nor Romney in terms of likeability = 7%

b) For a given survey of 500, the number of respondents that favored Obama than Romney is 145.

Step-by-step explanation:

Given that none of those surveyed can favour the two candidates at the same time,

n(Universal set) = n(U) = 100%

n(Obama) = n(O) = 61%

n(Romney) = n(R) = 32%

n(That favour Obama and Romney) = n(O n R) = 0%

To calculate for the number that favour neither of the candidates

n(O' n R')

n(U) = n(O) + n(R) + n(O n R) + n(O' n R')

100 = 61 + 32 + 0 + n(O' n R')

n(O' n R') = 100 - 93 = 7%

b) For a given survey of 500, how many more respondents favored Obama than Romney?

Number of those surveyed that favour Obama = 61% of 500 = 305

Number of those surveyed that favour Romney = 32% of 500 = 160

Difference = 305 - 160 = 145

5 0

1 year ago

Of 1,050 randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wage ea

garik1379 [7]

Answer:

We can claim with 95% confidence that the proportion of executives that prefer trucks is between 19.2% and 32.8%.

Step-by-step explanation:

We have a sample of executives, of size n=160, and the proportion that prefer trucks is 26%.

We have to calculate a 95% confidence interval for the proportion.

The sample proportion is p=0.26.

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.26*0.74}{160}}\\\\\\ \sigma_p=\sqrt{0.0012}=0.0347$

The critical z-value for a 95% confidence interval is z=1.96.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_p=1.96 \cdot 0.0347=0.068$

Then, the lower and upper bounds of the confidence interval are:

$LL=p-z \cdot \sigma_p = 0.26-0.068=0.192\\\\UL=p+z \cdot \sigma_p = 0.26+0.068=0.328$

The 95% confidence interval for the population proportion is (0.192, 0.328).

We can claim with 95% confidence that the proportion of executives that prefer trucks is between 19.2% and 32.8%.

3 0

2 years ago

A random sample of size 100 was taken from a population. A 94% confidence interval to estimate the mean of the population was co

Firlakuza [10]

Answer:

Step-by-step explanation:

1) The z value was determined using a normal distribution table. From the normal distribution table, the corresponding z value for a 94% confidence interval is 1.88

The correct option is D

2) if 70% of all internet users experience e-mail fraud. It means that probability of success, p

p = 70/100 = 0.7

q = 1 - p = 1 - 0.7 = 0.3

n = number of selected users = 50

Mean, u = np = 50×0.7 = 35

Standard deviation, u = √npq = √50×0.7×0.3 = 3.24

x = number of internet users

The formula for normal distribution is expressed as

z = (x - u)/s

We want to determine the probability that no more than 25 were victims of e-mail fraud. It is expressed as

P(x lesser than or equal to 25)

The z value will be

z = (25- 35)/3.24 = - 10/3.24 = -3.09

Looking at the normal distribution table, the corresponding z score is 0.001

P(x lesser than or equal to 25) = 0.001

7 0

2 years ago

A web-based company has a goal of processing 90 percent of its orders on the same day they are received. If 434 out of the next

Kamila [148]

Answer:

We conclude that a web-based company are not exceeding their goal of 90%.

Step-by-step explanation:

We are given that a web-based company has a goal of processing 90 percent of its orders on the same day they are received.

434 out of the next 471 orders are processed on the same day.

Let p = proportion of orders processing on the same day they are received.

SO, Null Hypothesis, $H_0$ : p $\leq$ 0.90 {means that they are not exceeding their goal of 90%}

Alternate Hypothesis, $H_0$ : p > 0.90 {means that they are exceeding their goal of 90%}

The test statistics that would be used here One-sample z test for proportions;

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

where, $\hat p$ = sample proportion of orders that are processed on the same day = $\frac{434}{471}$ = 0.92

n = sample of orders = 471

So, the test statistics = $\frac{0.92-0.90}{\sqrt{\frac{0.92(1-0.92)}{471} } }$

= 1.599

The value of z test statistics is 1.599.

Now, at 0.025 significance level the z table gives critical value of 1.96 for right-tailed test.

Since our test statistic is less than the critical value of z as 1.599 < 1.96, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that a web-based company are not exceeding their goal of 90%.

8 0

1 year ago

Robby had 4 4/9 bags of pet food.All of the bags held the same amount of food when they were full.Of Robby's 4 4/9 bags of pet f

Tpy6a [65]

Answer: 1.8 bags

Step-by-step explanation:

From the question, given that robby have 4 4/9 bags of pet food, 2/5 were dog food

2/5 of 4 4/9 = dog foods

Convert 4 4/9 to proper fraction= 40/9

2/5 of 40/9 means 2/5 × 40/9

= 16/9

= 1.78 or 1.8

I hope this helps, please mark as brainliest answer.

4 0

1 year ago