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3 years ago

Which statement is true about 8 in the whole numbers 1,825 and 8,367

Mathematics

2 answers:

Marat540 [252]3 years ago

4 0

The 8 is in the hundred spot in 1825 and in the thousand spot in 8367

Alenkinab [10]3 years ago

3 0

The eight in 8367 is 100 more than the eight in 1825

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Please help<br> Regroup and write your answer in simplest form

garik1379 [7]

First you should write 2 3/4 as a decimal, 2.75 then multiply 2.75 by 2 to get 5.5. 5.5 equals 5 1/2.
Ms. Addison needs 5 1/2 cups of flour.

7 0

3 years ago

Factor tree<br>67344 using prime numbers

Aleksandr-060686 [28]

Answer:

$2^4*3*23*61$

Step-by-step explanation:

$\begin{array}{c|c}67344&2\\33672&2\\16836&2\\8418&2\\4209&3\\1403&23\\61&61\\1&\\\end{array}$

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3 years ago

The graph of a linear function passes through the points (2, 4) and (8, 10).

Keith_Richards [23]

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 2}}\quad ,&{{ 4}})\quad 
% (c,d)
&({{ 8}}\quad ,&{{ 10}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{10-4}{8-2}$

$\bf \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\qquad 
\begin{array}{llll}
\textit{plug in the values for }
\begin{cases}
y_1=4\\
x_1=2\\
m=\boxed{?}
\end{cases}\\
\textit{and solve for "y"}
\end{array}$

8 0

3 years ago

A distribution of values is normal with a mean of 188 and a standard deviation of 20.8. Find the probability that a randomly sel

padilas [110]

Answer:0.8413

Step-by-step explanation:

Mean= 188, Std Dev. =20.8, Z=(x-mean)/Std Dev

P(X less than 167.2)= P(Z less than (167.2-188)/20.8)

=P(Z less than (-20.8/20.8))

=P(z less than -1)= P(Z less 1)

The value of 1 in the normal distribution table is 0.3413

So we add 0.5 to 0.3413 =0.8413

6 0

2 years ago

A toy manufacturer wants to know how many new toys children buy each year. A sample of 601 children was taken to study their pur

Lilit [14]

Answer:

The confidence interval is 6.6<μ<6.8.

Step-by-step explanation:

We have:

Number of observations = 601

Mean = 6.7

Standard deviation σ = 1.5

The z-score for a 95% confidence interval is 1.96.

The limits of the confidence interval can be calculated as

$X \pm z*\frac{\sigma}{\sqrt{n}}\\\\LL=X-z*\frac{\sigma}{\sqrt{n}}=6.7-1.96*\frac{1.5}{\sqrt{601} } =6.7-0.1199=6.6\\\\UL=X+z*\frac{\sigma}{\sqrt{n}}=6.7+1.96*\frac{1.5}{\sqrt{601} } =6.7+0.1199=6.8$

The confidence interval is 6.6<μ<6.8.

7 0

3 years ago