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Ivahew [28]
2 years ago
7

Help! Pls answer correctly!​

Mathematics
1 answer:
defon2 years ago
8 0

Answer:

<h3>numbers of Ostrich he keep = 725</h3>

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

  • Area of field = 30 × 155 + 90 × 45
  • Area of field = 4650 + 4050
  • Area of field = 8700 m²

  • 1 ostrich need = 12 m²

numbers of Ostrich = 8700 ÷ 12

numbers of Ostrich = 725

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Simplify the radicals: do 10- 18, no need to show work if you usually do, thank you!!
Ainat [17]

Answer:

10. 7

11. 4

12. 8

13. 7

14. 2

15. 3

16. 11

17. 9

18. 8

Step-by-step explanation:

3 0
3 years ago
A school is selling tickets for the prom. Tickets with a picture package cost $50 and tickets without the packet cost $30. If th
krek1111 [17]

Answer:

Answer

1.0/5

0

1 answer

2780/20 = 139 picture tickets

175-139 =36other tickets

43 - 18*2 = 43 -36 = 7

7 were 3 point shots

11 were 2 point shots

Step-by-step explanation:

8 0
3 years ago
Joe is training for a race in 30 days. He needs to run a total of 135 miles. How many miles will joe need to run each day to be
marin [14]
He needs to run 4.5 miles each day.
4 0
3 years ago
An engineer deposits $2,000 per month for four years at a rate of 24% per year, compounded semi-annually. How much will he be ab
Alex17521 [72]

The amount of money he will be able to withdraw after 10 years after his last deposit is $926,400.

<h3>Compound interest</h3>

  • Principal, P = $2,000 × 12 × 4

= $96,000

  • Time, t = 10 years
  • Interest rate, r = 24% = 0.24
  • Number of periods, n = 2

A = P(1 + r/n)^nt

= $96,000( 1 + 0.24/2)^(2×10)

= 96,000 (1 + 0.12)^20

= 96,000(1.12)^20

= 96,000(9.65)

= $926,400

Therefore, the amount of money he will be able to withdraw after 10 years after his last deposit is $926,400

Learn more about compound interest:

brainly.com/question/24924853

#SPJ4

5 0
2 years ago
While conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modem
Kitty [74]

Answer:

We conclude that this is an unusually high number of faulty modems.

Step-by-step explanation:

We are given that while conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modems.

The probability of obtaining this many bad modems (or more), under the assumptions of typical manufacturing flaws would be 0.013.

Let p = <em><u>population proportion</u></em>.

So, Null Hypothesis, H_0 : p = 0.013      {means that this is an unusually 0.013 proportion of faulty modems}

Alternate Hypothesis, H_A : p > 0.013      {means that this is an unusually high number of faulty modems}

The test statistics that would be used here <u>One-sample z-test</u> for proportions;

                             T.S. =  \frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }  ~  N(0,1)

where, \hat p = sample proportion faulty modems= \frac{10}{367} = 0.027

           n = sample of modems = 367

So, <u><em>the test statistics</em></u>  =  \frac{0.027-0.013}{\sqrt{\frac{0.013(1-0.013)}{367} } }

                                     =  2.367

The value of z-test statistics is 2.367.

Since, we are not given with the level of significance so we assume it to be 5%. <u>Now at 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.</u>

Since our test statistics is more than the critical value of z as 2.367 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.

Therefore, we conclude that this is an unusually high number of faulty modems.

6 0
3 years ago
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