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3 years ago

8

Please help asap!!! What are the image points

1 answer:

DerKrebs [107]3 years ago

3 0

Bbshehshsjehebnnejenwnsbwnsnsnsne

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A baker mixes a 21-pound batch of dough that contains 3 pounds of sugar. Approximately how many pounds of sugar must be added fo

scZoUnD [109]

The answer is c. 3.6

3.6 + 3 = 6.6 pounds of sugar
6.6-3 = 3.6 pounds of sugar added to the dough. This weight that must be added to the total weight of the dough.
3.6 (new sugar) +21 (original dough weight) = 24.6 (total weight)

6.6/24.6 = 26.8%

7 0

3 years ago

PLZ HELP GUYS............

vitfil [10]

Answer:

11

Step-by-step explanation:

10+1=11 lol bruh

4 0

3 years ago

Clara lends half her collection of formal attires to her sister, Susan. Clara then buys four more attires. If she has 12 attires

nirvana33 [79]

Answer:

i think 4 attires

Step-by-step explanation:

5 0

3 years ago

Suppose a random variable x is best described by a uniform probability distribution with range 22 to 55. Find the value of a tha

const2013 [10]

Answer:

(a) The value of <em>a</em> is 53.35.

(b) The value of <em>a</em> is 38.17.

(c) The value of <em>a</em> is 26.95.

(d) The value of <em>a</em> is 25.63.

(e) The value of <em>a</em> is 12.06.

Step-by-step explanation:

The probability density function of <em>X</em> is:

$f_{X}(x)=\frac{1}{55-22}=\frac{1}{33}$

Here, 22 < X < 55.

(a)

Compute the value of <em>a</em> as follows:

$P(X\leq a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.95\times 33=[x]^{a}_{22}\\\\31.35=a-22\\\\a=31.35+22\\\\a=53.35$

Thus, the value of <em>a</em> is 53.35.

(b)

Compute the value of <em>a</em> as follows:

$P(X< a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.49\times 33=[x]^{a}_{22}\\\\16.17=a-22\\\\a=16.17+22\\\\a=38.17$

Thus, the value of <em>a</em> is 38.17.

(c)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.85=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.85\times 33=[x]^{55}_{a}\\\\28.05=55-a\\\\a=55-28.05\\\\a=26.95$

Thus, the value of <em>a</em> is 26.95.

(d)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.89=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.89\times 33=[x]^{55}_{a}\\\\29.37=55-a\\\\a=55-29.37\\\\a=25.63$

Thus, the value of <em>a</em> is 25.63.

(e)

Compute the value of <em>a</em> as follows:

$P(1.83\leq X\leq a)=\int\limits^{a}_{1.83} {\frac{1}{33}} \, dx \\\\0.31=\frac{1}{33}\cdot \int\limits^{a}_{1.83} {1} \, dx \\\\0.31\times 33=[x]^{a}_{1.83}\\\\10.23=a-1.83\\\\a=10.23+1.83\\\\a=12.06$

Thus, the value of <em>a</em> is 12.06.

7 0

3 years ago

Which expression is equivalent to (7 – 3k) – (4 – 5k) – (6 – 2k)?

gladu [14]

Answer:

The right response is " $3(k-1)$ ". A further solution is given below.

Step-by-step explanation:

The given expression is:

⇒ $(7 -3k) - (4 - 5k) - (6 - 2k)$

On solving the above expression, we get

⇒ $7-3k-4+5k-6+2k$

⇒ $7-3k-10+7k$

⇒ $3k-3$

On taking "3" as common, we get

⇒ $3(k -1 )$

3 0

3 years ago