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2 years ago

What is 15 mL to the nearest 0.1 mL?

1 answer:

4 0

The answer is 20ml. You have 15 and round it up because if your number in the ones digit is 5 or higher you round up. In this case you would round 15ml up to 20ml.

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James sets off from home on his bike.The graph shows his journey.A)what is James average speed on his outward journey?B)How long

katen-ka-za [31]

Answer:

a) ~10.58 km/h

b) 45 min

c) 15 km/h

Step-by-step explanation:

a) 45km for 4 hours and 15min (4.25 hours) = 45 / 45.25 ~= 10.58 km/h

b) From 4:15 to 5 = 45min

c) 45km for 3 hours (5:00 to 8:00) = 45 / 3 = 15 km/h

NOTE!

The author actually wanted the answer on A) to be 10 km/h and the journey to be 4.5 hours (4 hours 30min) but instead they made it 4.25 (as shown on the graph, since 1 square is 30 minutes).

5 0

3 years ago

Help ASAP can u explain a little too? (pls help if u don’t know it don’t answer pls) Thank You:)

larisa [96]

Since they are similar, the dimensions are in the same ratio. L1 = 5, L2 = 15, so they are in a 3:1 ratio. So if V1 = 60, then W1×H1 = 60/5 = 12
W2 must also be 3×W1 and H2 3×H1, and
3×3 = 9. So take 12×9 (W×H1×9) ×15 (L2) = V2
V2 = 12×9×15 = 1620 cm^3

Let me know the right answer when you find out!

8 0

3 years ago

Rewa Delta Union Rugby CEO has become concerned about the slow pace of the rugby games played in the current union rugby, fearin

ozzi

Answer:

We conclude that the mean duration of 15-sided union rugby games has decreased after the meeting.

Step-by-step explanation:

We are given that Before the meeting, the mean duration of the 15-sided rugby game time was 3 hours, 5 minutes, that is, 185 minutes.

A random sample of 36 of the 15-sided rugby games after the meeting showed a mean of 179 minutes with a standard deviation of 12 minutes.

Let $\mu$ = mean duration of 15-sided union rugby games after the meeting.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 185 minutes {means that the mean duration of 15-sided union rugby games has increased or remained same after the meeting}

Alternate Hypothesis, $H_A$ : $\mu$ < 185 minutes {means that the mean duration of 15-sided union rugby games has decreased after the meeting}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean duration of 15-sided union rugby games = 179 min

s = sample standard deviation = 12 minutes

n = sample of 15-sided rugby games = 36

So, the test statistics = $\frac{179-185}{\frac{12}{\sqrt{36} } }$ ~ $t_3_5$

= -3

The value of t test statistics is -3.

Now, at 1% significance level the t table gives critical value of -2.437 at 35 degree of freedom for left-tailed test.

Since our test statistic is less than the critical value of t as -3 < -2.437, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean duration of 15-sided union rugby games has decreased after the meeting.

6 0

3 years ago

Please help me get this done. due at 11:59 tonight

kakasveta [241]

Answer:

m∠C = 30°

Step-by-step explanation:

If ΔADB is an equilateral, then m∠A = m∠ADB = m∠DBA = 60°

If ΔDBC isosceles with DB ≅ BC, then m∠C = m∠BDC ;

m∠C + m∠BDC = m∠DBA = 60° ⇒ m∠C = 30°

6 0

2 years ago

Write two subtraction equations to show 15-7.

kirill [66]

15 - 7 = 8
15 - 7 - 0 = 8

7 0

3 years ago