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nexus9112 [7]
3 years ago
14

Simplify: (7x4 – 7x3 + 7x2 + 2x) + (-5x4 + 7x3 + 5x2 – 9)​

Mathematics
1 answer:
ikadub [295]3 years ago
8 0

Answer:

Step-by-step explanation:

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Jasmine is baking and selling cupcakes for $15 per dozen she would like to make more than $120 in cupcake sales which number lin
antoniya [11.8K]

Answer:

 i'll explain it to you

Step-by-step explanation: look for a number line that has 180, 120, and 15 you see you have to multiply 15 by 12 dozen cupcakes which equals to 180  that way she earns more than $120

3 0
3 years ago
An electronics store received two shipments of radios. Of the 250 radios in the first shipment, exactly 6% were damaged. In the
nataly862011 [7]

Answer:

Total number of not damaged = 451

C. 451

Step-by-step explanation:

In the first shipment, 6% of the 250 radios where damaged.

Number of damaged= 6/100 * 250

Number of damaged= 6*2.5

Number of damaged= 15 radios.

Number of not damaged= 250-15

Number of not damaged = 235 radios

In the second shipment, 4% of the radios were damaged

Let the number of second shipment= y

0.04 of y where damaged

Total number of radio damaged= 24

15 + 0.04y = 24

0.04y= 24-15

0.04y = 9

Y= 9/0.04

Y= 225

Number of radio in the second shipment= 225

Number of damaged in the second shipment= 0.04*225

Number of damaged in the second shipment= 9

Number of not damaged in the second shipment= 225-9

Number of damaged in the second shipment= 216

Total number of not damaged

= 235+216

= 451

Total number of not damaged= 451

7 0
3 years ago
Help please...................
lys-0071 [83]

Answer:

ok so what im seeing is a mickey mouse on the paper, the answer to all question has to mikey mouse

Step-by-step explanation:

mickey mouse is on the paper like its giving you the answer? why would you need help ITS SO SIMPLE

8 0
3 years ago
If the angles of a triangle are (2y - 5)degrees, (y + 20)degrees, and (3y-5) degrees what are the values of the angles​
NNADVOKAT [17]

Answer:

<h3>\boxed{ \boxed{ \bold{ \sf{51.68  \: , \: 48.34 \:,  \: 80.02}}}}</h3>

Step-by-step explanation:

Let's solve:

As we know that the sum of angles of traingle adds to 180°

\sf{2y - 5 + y + 20 + 3y - 5 = 180}

Collect like terms

⇒\sf{6y - 5 + 20 - 5 = 180}

⇒\sf{6y   + 15 - 5 = 180}

⇒\sf{6y + 10 = 180}

Move constant to right hand side and change it's sign

⇒\sf{6y = 180 - 10}

Calculate the difference

⇒\sf{6y = 170}

Divide both sides of the equation by 6

⇒\sf{ \frac{6y}{6}  =  \frac{170}{6} }

Calculate

⇒\sf{y = 28.34}

Now, let's replace the value:

⇒\sf{2y - 5 = 2 \times 28.34 - 5 = 51.68}

⇒\sf{y + 20 = 28.34  + 20 = 48.34}

⇒\sf{3y - 5 = 3 \times 28.34 - 5 = 80.02}

Hope I helped!

Best regards!!

8 0
3 years ago
Read 2 more answers
A random sample of 500 registered voters in Phoenix is asked if they favor the use of oxygenated fuels year-round to reduce air
Stells [14]

Answer:

a) 0.0853

b) 0.0000

Step-by-step explanation:

Parameters given stated that;

H₀ : <em>p = </em>0.6

H₁ : <em>p  = </em>0.6, this explains the acceptance region as;

p° ≤ \frac{315}{500}=0.63 and the region region as p°>0.63 (where p° is known as the sample proportion)

a).

the probability of type I error if exactly 60% is calculated as :

∝ = P (Reject H₀ | H₀ is true)

   = P (p°>0.63 | p=0.6)

where p° is represented as <em>pI</em><em> </em>in the subsequent calculated steps below

   

    = P  [\frac{p°-p}{\sqrt{\frac{p(1-p)}{n}}} >\frac{0.63-p}{\sqrt{\frac{p(1-p)}{n}}} |p=0.6]

    = P  [\frac{p°-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} >\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} ]

    = P   [Z>\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500} } } ]

    = P   [Z > 1.37]

    = 1 - P   [Z ≤ 1.37]

    = 1 - Ф (1.37)

    = 1 - 0.914657 ( from Cumulative Standard Normal Distribution Table)

    ≅ 0.0853

b)

The probability of Type II error β is stated as:

β = P (Accept H₀ | H₁ is true)

  = P [p° ≤ 0.63 | p = 0.75]

where p° is represented as <em>pI</em><em> </em>in the subsequent calculated steps below

  = P [\frac{p°-p} \sqrt{\frac{p(1-p)}{n} } }\leq \frac{0.63-p}{\sqrt{\frac{p(1-p)}{n} } } | p=0.75]

  = P [\frac{p°-0.6} \sqrt{\frac{0.75(1-0.75)}{500} } }\leq \frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]

  = P[Z\leq\frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]

  = P [Z ≤ -6.20]

  = Ф (-6.20)

  ≅ 0.0000 (from Cumulative Standard Normal Distribution Table).

6 0
3 years ago
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