<img src="https://tex.z-dn.net/?f=-2b-%28-7%29%3D11" id="TexFormula1" title="-2b-(-7)=11" alt="-2b-(-7)=11" align="absmiddle" cl

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maks197457 [2]

3 years ago

ass="latex-formula">

Mathematics

2 answers:

Agata [3.3K]3 years ago

7 0

Answer:

b = -2

Step-by-step explanation:

$- 2b - ( - 7) = 11$

$- 2b + 7 = 11$

$- 2b + 7 - 7 = 11 - 7$

$- 2b = 11 - 7$

$- 2b = 4$

$b = 4 \div( - 2) = - 4 \div 2$

$\boxed{\green{b = - 2}}$

Fittoniya [83]3 years ago

5 0

<h2>Answer:</h2>

<h2>Step-by-step explanation:</h2>

$- 2b - ( - 7) = 11 \\ \\ - 2b + 7 = 11 \\ \\ - 2b = 11 - 7 \\ \\ - 2b = 4 \\ \\ b = \frac{4}{ 2} \\ \\ b = 2$

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Y= 99 9. Suppose a varies directly as b, and a = 7 when b = 2. Find b when a = 21.

kogti [31]

Answer:

Step-by-step explanation:

Given expression:

a varies directly as b;

a ∝ b

So;

a = kb where k is the constant of proportionality;

if a= 7 and b = 2;

7 = 2k

k = $\frac{7}{2}$

Now, find b when a = 21

21 = $\frac{7}{2}$ x b

b = 6

5 0

3 years ago

A bag contains purple marbles and white marbles, 96 in total. The number of purple marbles is 6 less than 5 times the number of

love history [14]

PURPLE MARBLES=P
WHITE MARBLES=W

First, write out an equation showing the number of purple marbles.
P=5W-6
Then, write out the main equation.
W+P=96
W+5W-6=96
Add 6 to both sides.
W+5W=102
6W=102
Divide both sides by 6
W=17
Now that we know there are 17 white marbles, we can go back to the equation about how many purple marbles there are.
P=5W-6
Plug in the value for W, which is 17.
P=5(17)-6
P=85-6
P=79

To make sure your answer is correct, check your work by plugging in the values for P+W=96.
79+17=96
Answer: There are 79 purple marbles.

8 0

3 years ago

The ideal (daytime) noise-level for hospitals is 45 decibels with a standard deviation of 10 db; which is to say, this may not b

WINSTONCH [101]

Answer:

(a) A 99% confidence interval for the actual mean noise level in hospitals is (44.02 db, 49.98 db).

(b) We can be 90% confident that the actual mean noise level in hospitals is 47 db with a margin of error of 1.89 db.

(c) Unless our sample (of 81 hospitals) is among the most unusual 2% of samples, the actual mean noise level in hospitals is between 44.41 db and 49.59 db.

Step-by-step explanation:

The problem is incomplete. The questions are:

(a) A 99% confidence interval for the actual mean noise level in hospitals is (44.02 db, 49.98 db).

For a 99% CI, the value of z is z=2.58

Then, the confidence interval for the mean is:

$M-z\sigma/\sqrt{n}\leq\mu\leq M-z\sigma/\sqrt{n}\\\\47-2.58*10/\sqrt{75} \leq\mu\leq47+2.58*10/\sqrt{75}\\\\47-2.98\leq\mu\leq47+2.98\\\\44.02\leq\mu\leq 49.98$

(b) We can be 90% confident that the actual mean noise level in hospitals is 47 db with a margin of error of 1.89 db.

For a 90% CI, the value of z is z=1.64.

Then, we can calculate the margin of error as:

$E=z*\sigma/\sqrt{n}=1.64*10/\sqrt{75}=1.89$

(c) Unless our sample (of 81 hospitals) is among the most unusual 2% of samples, the actual mean noise level in hospitals is between 44.41 db and 49.59 db.

The 2% tails data corresponds, in the standard normal distirbution, to the values of z whose absolute value is higher than 2.33.

The values of db for these critical values are:

$X_1=M+z_1*\sigma/\sqrt{n}=47+(-2.33)*10/\sqrt{81}=47-2.59=44.41\\\\\\ X_2=M+z_2*\sigma/\sqrt{n}=47+(2.33)*10/\sqrt{81}=47+2.59=49.59$