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2 years ago

10

What term refers to common, non-rechargeable batteries?

1 answer:

kiruha [24]2 years ago

6 0

Answer:

Lithium

Explanation:

Lithium batteries features primary cell construction.They are single-use or non rechargeable batteries

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If Manuel does a job in 150 hours and with the help of Shantel they can do it together in 50 hours, how long would it take Shant

Alex Ar [27]

75 hours because 150 divided by two is 75

6 0

3 years ago

Minnesota Public Radio wants to duplicate a survey conducted in 2016 that found that 65% of adults living in Minnesota felt that

vesna_86 [32]

Answer:

The number of people needed is $n =684$

Step-by-step explanation:

From the question we are told that

The population proportion is $p = 0.65$

The margin of error is $E = 0.03$

From the question we are told the confidence level is 90% , hence the level of significance is

$\alpha = (100 - 90 ) \%$

=> $\alpha = 0.10$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the sample size is mathematically represented as

$n =[ \frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * p(1-p)$

=> $n =[ \frac{1.645 }{0.03} ]^2 * 0.65(1-0.65)$

=> $n =684$

6 0

3 years ago

Anna baked 3 batches of cookies with c cookies in each batch. She then ate 8 cookies! How many cookies does Anna have left? Writ

borishaifa [10]

Answer:

$3c-8$

Step-by-step explanation:

Let

c-------> the number of cookies in each batch

we know that

To find the total number of cookies, multiply the number of batches by the number of cookies in each batch

so

$3c$ ------> equation that represent the total number of cookies

If Anna ate $8$ cookies

then

The expression to know how many cookies Anna has left is equal to

$3c-8$

8 0

3 years ago

g The manager of a grocery store has selected a random sample of 100 customers. The average length of time it took these 100 cus

vagabundo [1.1K]

Answer:

The right solution is "0.196".

Step-by-step explanation:

The given values are:

Standard deviation,

$\sigma=1 \ minute$

Sample size,

$n=100$

Confidence level,

$c=95$ %

$=0.95$

Now,

For $\alpha=1-c$

$= 1-0.95$

$=0.05$

then,

$Z_{\frac{\alpha}{2} } = Z_{\frac{0.05}{2} }$

$=Z_{0.025}$

$=1.96$

The margin or error will be:

⇒ $E=Z_{\frac{\alpha}{2} } (\frac{\sigma}{\sqrt{n} })$

On substituting the values, we get

= $1.96(\frac{1}{\sqrt{100} })$

= $1.96\times \frac{1}{10}$

= $0.196$

7 0

3 years ago

Can I get help please!

Naily [24]

55x5 - 40x5
The third one down

8 0

3 years ago