-9/11 multiplied by -11/9

What is the value of x if 5x+2=5^9 х = -11 X=-7 x=7 X=11

Lemur [1.5K]

X=7 7 is the value of this math problem

6 0

3 years ago

The sale price of every item in a store is 85% of its usual price.

nydimaria [60]

Answer:

the sales price of backpack = $25.3

the sale price of sweatshirt = $15.3

The sale price of soccer ball = $24.80

8 0

3 years ago

In a survey of 1,003 adults concerning complaints about restaurants, 732 complained about dirty or ill-equipped bathrooms and 38

Ganezh [65]

Answer:

a

$0.716 < p < 0.744$

b

$0.3498 < p < 0.4089$

c

With the result obtained from a and b the manager can be 95 % confidence that the proportion of the population that complained about dirty or ill-equipped bathrooms are within the interval obtained at a

and that

the proportion of the population that complained about loud or distracting diners at other tables are within the interval obtained at b

Step-by-step explanation:

From the question we are told that

The sample size is $n = 1003$

The number that complained about dirty or ill-equipped bathrooms is $e = 732$

The number that complained about loud or distracting diners at other tables is $q = 381$

Given that the the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = (100- 95)\%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table , the value is

$Z_{\frac{\alpha }{2} } = 1.96$

Considering question a

The sample proportion is mathematically represented as

$\r p = \frac{e}{n}$

=> $\r p = \frac{732}{1003}$

=> $\r p = 0.73$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{ \r p (1- \r p)}{n} }$

$E = 1.96* \sqrt{ \frac{ 0.73 (1- 0.73)}{1003} }$

$E = 0.01402$

The 95% confidence interval is

$\r p - E < p < \r p +E$

$0.73 - 0.01402 < p < 0.73 + 0.01402$

$0.716 < p < 0.744$

Considering question b

The sample proportion is mathematically represented as

$\r p = \frac{q}{n}$

=> $\r p = \frac{381}{1003}$

=> $\r p = 0.3799$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{ \r p (1- \r p)}{n} }$

$E = 1.96* \sqrt{ \frac{ 0.3799 (1- 0.3799)}{1003} }$

$E = 0.0300$

The 95% confidence interval is

$\r p - E < p < \r p +E$

$0.3798 - 0.0300 < p < 0.3798 + 0.0300$

$0.3498 < p < 0.4089$

8 0

3 years ago

Juan received scores of 82, 76, 93, and 80 on his first four chemistry tests of the year. His goal is to have an 86 average in c

SCORPION-xisa [38]

Answer:

99 marks

Step-by-step explanation:

Given:

Juan received scores of 82, 76, 93, and 80 on his first four chemistry tests.

His goal is to have an 86 average in chemistry for his first five tests.

Question asked:

What score must he earn on the next test to achieve an average of exactly 86?

Solution:

Let score he must earn on the fifth test = $x$

As we know:

Scores in five test = 82, 76, 93, 80, $x$

Total number of test = 5

We can find the score of fifth test, he must earn to achieve an average of exactly 86 by using this formula:

$Average =\frac{Sum\ of\ all\ observations}{Total\ observation}$

$86=\frac{82+76+93+80+x}{5}\\\\ 86=\frac{331+x}{5}$

By cross multiplication

$331+x=86\times5\\331+x=430$

Subtracting both sides by 331

$x=99$

Therefore, he must earn 99 marks to achieve an average of exactly 86 for his first five tests in chemistry.

6 0

3 years ago

How do you figure out what numbers are terminating and what numbers are repeating ?

raketka [301]

You can simply look and see if one is terminating by seeing if it ends. For example, 2.54 is terminating since the numbers after the decimal aren't repeating. A repeating number is for example 0.555555..

3 0

3 years ago