All categories

3 years ago

18÷3(5-4+1) =?? 3 or 12 how ?!

Mathematics

2 answers:

larisa86 [58]3 years ago

7 0

Answer: answer is 3

Step-by-step explanation:

18 ÷ 3(5-4+1)

Step 1: apply BODMAS

Open bracket first, then add before subtracting

= 18÷ 3 × 1+1

18÷3 x 2

18 ÷ 6

sergejj [24]3 years ago

3 0

Answer:

Step-by-step explanation:

Parenthesis first:

5-4+1

5-4=1+1=2

18÷3×2

Divide:

18÷3×2

Multiply:

6×2=12

You might be interested in

15 points. Need help with this math question

kodGreya [7K]

Answer:

0.0098 < 0.01935 < 0.02 < 0.14999 < 0.1589

8 0

3 years ago

Read 2 more answers

What is 6,007,200 in standard form

Orlov [11]

(6,000,000) (7000) (200)

6 0

3 years ago

Compare the sets of numbers given and place them in the correct column or incorrect column.

TEA [102]

Answer:

I think the answer is option 3

3 0

4 years ago

A sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.

TEA [102]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ = $1,150

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ $1,150

(b) The test statistic is 3.571.

(c) We conclude that the mean of all account balances is significantly different from $1,150.

Step-by-step explanation:

We are given that a sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.

We have test the hypothesis to determine whether the mean of all account balances is significantly different from $1,150.

Let $\mu$ = mean of all account balances

(a)So, Null Hypothesis, $H_0$ : $\mu$ = $1,150 {means that the mean of all account balances is equal to $1,150}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ $1,150 {means that the mean of all account balances is significantly different from $1,150}

The test statistics that will be used here is One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average balance = $1,200

s = sample standard deviation = $126

n = sample of account balances = 81

(b) So, test statistics = $\frac{1,200-1,150}{\frac{126}{\sqrt{81} } }$ ~ $t_8_0$

= 3.571

The value of the sample test statistics is 3.571.

(c) Now, P-value of the test statistics is given by the following formula;

P-value = P( $t_8_0$ > 3.571) = Less than 0.05%

Because in the t table the highest critical value for t at 80 degree of freedom is given between 3.460 and 3.373 at 0.05% level.

Now, since P-value is less than the level of significance as 5% > 0.05%, so we sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the mean of all account balances is significantly different from $1,150.

8 0

4 years ago

HELPPPPPPPPPPPPPPPPWhich statements are true about the ordered pair(-1, -4) and the system of

Anastasy [175]

Answer:

When (-1,-4) is substituted into the second equation, the equation is true.

The ordered pair (-1,-4) is a solution to the system of linear equations.

When (-1,-4) is substituted into the first equation, the equation is true.

Step-by-step explanation:

Given

$x - y = 3$

$7x - y = -3$

Required

What is true about $(x,y) = (-1,-4)$

To do this, we substitute -1 for x and -4 for y in both equations;

For $x - y = 3$

$-1 - (-4) = 3$

$-1 + 4 = 3$

$3 = 3$

Hence: $(-1,-4)$ is true for $x - y = 3$

For $7x - y = -3$

$7(-1) - (-4) = -3$

$-7 +4 = -3$

$-3 = -3$

Hence: $(-1,-4)$ is true for $7x - y = -3$

3 0

3 years ago