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Delvig [45]
3 years ago
5

Pls answer this with an explanation if you can pg3 pt1

Mathematics
2 answers:
SpyIntel [72]3 years ago
4 0

Answer:

A

Step-by-step explanation:

if he earned 9 dollars per hour, and 405 dollars total. You need to set up and equation. 9h=405. H stands for each hour. the equation is saying 9 times what is 405

can you please mark me brainliest (✿◠‿◠)

quester [9]3 years ago
3 0

Answer:

A

Step-by-step explanation:

You multiply the number of hours he worked to the amount he makes and set it equal to the total he made.

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What is the area of a circular region whose diameter is 10 centimeters?
stepan [7]

Answer:

78cm

Step-by-step explanation:

d = 10

area of a circle = πr^2

22/7 × 5 × 5

78.5cm

6 0
3 years ago
7x – 2y = -5 which is the correct answer help a homie out pls
Elis [28]
Neither of those work for that equation
7 0
3 years ago
1 Point
I am Lyosha [343]

Option C

The ratio for the volumes of two similar cylinders is 8 : 27

<h3><u>Solution:</u></h3>

Let there are two cylinder of heights "h" and "H"

Also radius to be "r" and "R"

\text { Volume of a cylinder }=\pi r^{2} h

Where π = 3.14 , r is the radius and h is the height

Now the ratio of their heights and radii is 2:3 .i.e  

\frac{\mathrm{r}}{R}=\frac{\mathrm{h}}{H}=\frac{2}{3}

<em><u>Ratio for the volumes of two cylinders</u></em>

\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{\pi r^{2} h}{\pi R^{2} H}

Cancelling the common terms, we get

\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\left(\frac{\mathrm{r}}{R}\right)^{2} \times\left(\frac{\mathrm{h}}{\mathrm{H}}\right)

Substituting we get,

\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\left(\frac{2}{3}\right)^{2} \times\left(\frac{2}{3}\right)

\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{2 \times 2 \times 2}{3 \times 3 \times 3}

\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{8}{27}

Hence, the ratio of volume of two cylinders is 8 : 27

7 0
3 years ago
The calibration of a scale is to be checked by weighing a 13 kg test specimen 25 times. Suppose that the results of different we
Natali5045456 [20]

Solution :

a).

Given : Number of times, n = 25

            Sigma, σ = 0.200 kg

            Weight, μ = 13 kg

Therefore the hypothesis should be tested are :

$H_0 : \mu = 13 $

$H_a : \mu \neq 13$

b). When the value of $\overline x = 12.84$

 Test statics :

   $Z=\frac{(\overline x - \mu)}{\frac{\sigma}{\sqrt n}} $

  $Z=\frac{(14.82-13)}{\frac{0.2}{\sqrt {25}}} $

          $=\frac{1.82}{0.04}$

          = 45.5  

P-value = 2 x P(Z > 45.5)

            = 2 x 1 -P (Z < 45.5) = 0

Reject the null hypothesis if P value  < α = 0.01 level of significance.

So reject the null hypothesis.

Therefore, we conclude that the true mean measured weight differs from 13 kg.

3 0
3 years ago
Simplify: 4w9+(-2w9) enter the original expression if it cannot be simplified
slavikrds [6]
18w (please check for sure though, cause im not 100% ) 
6 0
3 years ago
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