All categories

3 years ago

975 toffees are to be distributed among 23 children. Estimate the numbers of toffees that each child will get in nearest tens.

Mathematics

1 answer:

Aneli [31]3 years ago

3 0

Answer:

Step-by-step explanation:

Total number of toffees = 975

Total number of children = 23

Estimate the numbers of toffees that each child will get in nearest tens.

Number of toffees each child get = Total number of toffees / Total number of children

= 975 / 23

= 42.391304347826

Approximately,

Number of toffees each child get = 40 to the nearest tens

You might be interested in

How do i solve this equation

Gre4nikov [31]

X < -16 is the solution
Hope this helped

6 0

3 years ago

Based on the diagram above, which of the following does not make q and parallel

iragen [17]

C. Because it only gives the measure of angles on line p and not line q

3 0

2 years ago

(a) Select all that describe BD. B Perpendicular bisector of AC Angle bisector of LB Median of AABC Altitude of A ABC A None of

Fynjy0 [20]

a) Altitude of triangle ABC

b) Angle bisector of angle H

c) Median of triangle JKL

3 0

1 year ago

Compare these fractions to 1/2. less than 1/2 equal to 1/2 greater than 1/2.

spin [16.1K]

1. > = < = < >
2. 3/4
3. ? What do you mean by yes/no?

3 0

3 years ago

Read 2 more answers

The attention span of children (ages 3 to 5) is claimed to be Normally distributed with a mean of 15 minutes and a standard devi

Agata [3.3K]

Answer:

If the observed sample mean is greater than 17.07 minutes, then we would reject the null hypothesis.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 15 minutes

Sample size, n = 10

Alpha, α = 0.05

Population standard deviation, σ = 4 minutes

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 15\text{ minutes}\\H_A: \mu > 15\text{ minutes}$

Since the population standard deviation is given, we use one-tailed z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Now, $z_{critical} \text{ at 0.05 level of significance for one tail} = 1.64$

Thus, we would reject the null hypothesis if the z-statistic is greater than this critical value.

Thus, we can write:

$z_{stat} = \displaystyle\frac{\bar{x} - 15}{\frac{4}{\sqrt{10}} } > 1.64\\\\\bar{x} - 15>1.64\times \frac{4}{\sqrt{10}}\\\bar{x} - 15>2.07\\\bar{x} > 15+2.07\\\bar{x} > 17.07$

Thus, the decision rule would be if the observed sample mean is greater than 17.07 minutes, then we would reject the null hypothesis.

6 0

3 years ago