Ask question

Ask question

All categories

3 years ago

7

1. A town's low temperature was -8 degrees. The difference between the town's high and low temperature for January was 25 degree

s. What was the town's high temperature?

1 answer:

notsponge [240]3 years ago

5 0

The answer is 17

(-8)+25=17

(-8)+25

-8+25=17

You might be interested in

What is the following product

Ira Lisetskai [31]

Answer: The answer to your question will be -

5^√16x^4

7 0

3 years ago

Read 2 more answers

I need to figure out what x is

Svet_ta [14]

For all triangles, the angle sum is 180°, therfore we can add all the angles up to solve x
54 + 55 + x + 74 =180
183 + x =180
X = 180 - 183
X = - 3 °
Don't worry, negative result is normal and possible.

8 0

3 years ago

A chemist examines 12 sedimentary samples for bromide concentraction. The mean bromide concentration for the sample date is 0.43

Umnica [9.8K]

Answer:

a) $z = 1.645$

b) Lower endpoint: 0.422cc/m³

Upper endpoint: 0.452 cc/m³

Step-by-step explanation:

Population is approximately normal, so we can find the normal confidence interval.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.05 = 0.95$ , so $z = 1.645$ . This is the critical value, the answer for a).

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.645*\frac{0.0325}{\sqrt{12}} = 0.015$

The lower end of the interval is the sample mean subtracted by M. So it is 0.437 - 0.015 = 0.422cc/m³.

The upper end of the interval is the sample mean added to M. So it is 0.437 + 0.015 = 0.452 cc/m³.

b)

Lower endpoint: 0.422cc/m³

Upper endpoint: 0.452 cc/m³

4 0

3 years ago

larisa86 [58]

V=ca times height
1040=40 times h
H=1040 divide by 40
Height=26

3 0

3 years ago

Refer to the previous exercise. The researchers wanted a sufficiently large sample to be able to estimate the probability of pre

Maru [420]

Answer:

<em>The large sample n = 117.07</em>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that the estimate error (M.E) = 0.08

The proportion (p) = 0.75

q =1-p = 1- 0.75 =0.25

Level of significance = 0.05

Z₀.₀₅ = 1.96≅ 2

<u><em>Step(ii):-</em></u>

The Marginal error is determined by

M.E = $\frac{Z_{0.05} \sqrt{p(1-p)} }{\sqrt{n} }$

$0.08 = \frac{2 X \sqrt{0.75(1-0.75)} }{\sqrt{n} }$

Cross multiplication , we get

$\sqrt{n} = \frac{2 X \sqrt{0.75(1-0.75)} }{0.08 }$

√n = $\frac{2 X0.4330}{0.08} = 10.825$

squaring on both sides , we get

n = 117.07

<u><em>Final answer:-</em></u>

<em>The large sample n = 117.07</em>

6 0

2 years ago