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3 years ago

Write the phrase as an expression. Then evaluate when x=5. The sum of a number x and 4, all divided by 3

2 answers:

3 0

Sum of a number, x and 4 all divided by 3

sum of x and 4 means x+4
all divided by 3
$\frac{x+4}{3}$
evaluate for x=5

$\frac{5+4}{3}=$
$\frac{9}{3}=$
$3$

it's 3

lisabon 2012 [21]3 years ago

3 0

Answer:

STOP

Step-by-step explanation:

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Please help I don't understand this

Effectus [21]

The correct answer is: [C]:  " $\frac{7}{18}$ " .
____________________________________________________
Explanation:
____________________________________________________
Given:
____________________________________________________

2 / (9x) = 4 / 7 ; solve for "x" ;
____________________________________________________

Cross-multiply:

→ (9x)*4 = 2 * 7 ;

  → 36x = 14 ;

Divide each side of the equation by "36" ;
to isolate "x" on one side of the equation; & to solve for "x" ;

→ 36x / 36 = 14 / 36 ;

x = 14/ 36 ;

  → x = 14/36 = (14÷2) / (36÷2) = 7/18 ;
______________________________________________________

The answer is: " $\frac{7}{18}$ " ;

→ which is: Answer choice: [C]: " $\frac{7}{18}$ " .
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3 years ago

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Goryan [66]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

A simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute

Katyanochek1 [597]

Answer:

Step-by-step explanation:

Given that:

sample size n = 36

standard deviation = 10.1

level of significance ∝ = 0.10

The null hypothesis and the alternative hypothesis can be computed as follows:

$H_0 : \sigma = 10$

$H_1 : \sigma \neq 10$

The test statistics can be computed as follows:

$X^2 = \dfrac{(n -1)s^2 }{\sigma ^2}$

$X^2 = \dfrac{(36 -1)10.1^2 }{10^2}$

$X^2 = \dfrac{(35)102.01 }{100}$

$X^2 = \dfrac{3570.35 }{100}$

$X^2 =35.704$

degree of freedom = n - 1 = 36 - 1 = 35

Since this test is two tailed .

The P -value can be determined by using the EXCEL FUNCTION ( = 2 × CHIDIST(35.7035, 35)

P - value = 2 × 0.435163515

P - value = 0.8703 ( to four decimal places)

Decision Rule : To reject the null hypothesis if P - value is less than the 0.10

Conclusion: We fail to reject null hypothesis ( accept null hypothesis) since p-value is greater than 0.10 and we conclude that there is sufficient claim that the normal range of pulse rates of adults given as 60 to 100 beats per minute resulted to a standard deviation of 10 beats per minute.