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Tasya [4]
2 years ago
14

A shop has 55 posters for sale. 34 posters show sports. The rest of the posters show animals. How many posters show animals?

Mathematics
2 answers:
sdas [7]2 years ago
5 0

Answer: 21 posters is the answer.

maksim [4K]2 years ago
3 0
21 Posters.


55-34=21
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I need help with this math problem. Check images.
Artemon [7]

Answer:

1,2,4,6,8,9,10,13

Step-by-step explanation:

4 0
3 years ago
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maria brought a blouse on sale for 20% off. the sale price was $21.76. what was the original price? round to the nearest hundred
Degger [83]

Answer: $27.20

Step-by-step explanation:

She bought a blouse for 20% off so it means that she paid 80% of the original price.We can then set up the equation  

80% of x = 21.76   where x is the original price so solve for x

0.8x = 21.76  

x = 27.20

Which  means the original price is $27.20

4 0
3 years ago
Write the equation of a line that has a slope of -3 and goes through the point (5,7).
ss7ja [257]

Answer:

The equation of the line would be D) y - 7 = -3(x -5)

Step-by-step explanation:

In order to find this, we must first start with the base form of point-slope form.

y - y1 = m(x - x1)

Now we put the slope in for m and the point in for (x1, y1)

y - 7 = -3(x - 5)

6 0
3 years ago
The contents of a sample of 26 cans of apple juice showed a standard deviation of 0.06 ounces. We are interested in testing to d
NikAS [45]

Answer:

Option b. should not be rejected

Step-by-step explanation:

We are given that the contents of a sample of 26 cans of apple juice showed a standard deviation of 0.06 ounces.

We have to test whether the variance of the population is significantly more than 0.003, i.e.;

  Null Hypothesis, H_0 : \sigma = \sqrt{0.003}

Alternate Hypothesis, H_1 : \sigma > \sqrt{0.003}

The test statistics used here for testing variance is;

          T.S. = \frac{(n-1)s^{2} }{\sigma^{2} } ~ \chi^{2}__n_-_1

where, s = sample standard deviation = 0.06

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So, Test statistics = \frac{(26-1)0.06^{2} }{0.003 } ~ \chi^{2}__2_5

                            = 30

So, at 5% level of significance chi square table gives critical value of 37.65 at 25 degree of freedom. Since our test statistics is less than the critical so we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that null hypothesis should not be rejected and variance of population is 0.003.

5 0
3 years ago
Antonio’s weekly allowance is given by the equation A=0.5c+10, where c is the number of chores he does. If he received $16 in al
Vedmedyk [2.9K]
A=0.5(12)+10= 16. Antonio has 12 chores.
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2 years ago
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